you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag(s) contains the counterfeit coins with just _one_ weighing?
You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
Suppose a rectangle can be (in some way) entirely covered by 25 circular disks, each of radius 1. Can the same rectangle be entirely covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes?
When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
On April 14 2011 09:41 sidr wrote: For those who know a bit more math, I have a variation on #10. Suppose countably many prisoners (meaning we can assign each prisoner a natural number, ie there is a prisoner 1, prisoner 2, etc for all positive whole numbers) are given the following scenario: on the following day, an evil warden will assemble them in one (very large) room and give each of them a hat with a (not necessarily unique) natural number on it. Each prisoner will be able to see all hats except their own. Without any communication with other prisoners after receiving his or her hat, each prisoner will communicate with the warden what he or she thinks his or her number is. This communication occurs simultaneously; that is to say prisoner x has know knowledge of what prisoner y communicated to the warden (unless of course x=y...). The prisoners are all allowed free if and only if finitely many of them are wrong. Assume the prisoners know this will happen the following day, and are given a night to prepare a strategy. Is there a strategy which guarantees the prisoners all go free? Give the strategy or prove no such strategy exists.
Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
On April 10 2011 09:56 LastPrime wrote: http://www.sciencenews.org/view/generic/id/7649/title/Puzzling_Names_in_Boxes "Indeed, the probability that a random permutation of 2n objects has no cycle of length greater than n is at least 1 minus the natural logarithm of 2." Does anyone know how this probability is calculated?
On April 10 2011 05:16 cmpcmp wrote: You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
On April 10 2011 05:44 Joe12 wrote: A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Its a while since i heard this one, but im pretty sure the wording is correct..
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
On April 10 2011 07:14 MusicalPulse wrote: I like logic puzzles more than math puzzles so... These are my two favorites :D
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there? ----- The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
On April 10 2011 07:59 LastPrime wrote: A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
On April 10 2011 00:05 stepover12 wrote: 7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale? I'll add more when I can.
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
suppose you have a chess board with 2 opposite corners cut out as in picture there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
On April 10 2011 18:45 0x64 wrote: In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
On April 11 2011 06:34 MusicalPulse wrote: One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
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Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
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In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
EDIT: Single layer counts as an odd number btw.
Rephrase: Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
On April 11 2011 08:22 Zeroes wrote: Threads like these make me think that the starcraft community is the smartest community! =D
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
On April 11 2011 12:05 x-Catalyst wrote: 4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
On April 11 2011 13:12 Akari Takai wrote: Think I'll contribute my own as well....
Suppose you are being screened for a disease that affects 1 in 10,000 people statistically. You are to be given a test for the disease that is 99% sensitive and 99% specific, that is, that the test will correctly identify someone who has the disease as testing positive 99% of the time, and will correctly identify someone who does not have the disease as testing negative 99% of the time. How much more likely are you to have the disease given a positive test result than to not have the disease given a positive test result.
I'm going to give the answer below just because I might not be in this thread for a few days and it would be unfair of me to not give the answer! But you should try to solve it on your own first.
Let P(S) = the probability that you are sick (This is 0.0001 given that the incidence of disease is 1 in 10,000 people) P(N) = the probability that you are not sick (This is 1-P(N) or 0.9999) P(+|S) = the probability that the test is positive, given that you are sick (This is 0.99 since the test is 99% accurate) P(+|N) = the probability that the test is positive, given that you are not sick (This is 0.01 since the test will produce a false positive for 1% of users) P(+) = the probability of a positive test event, regardless of other information. (0.010098, which is found by adding the probability that a true positive result will appear (0.99*0.0001) plus the probability that a false positive will appear (0.01*0.9999).)
P(N|+) = the probability that a non sick person tested positive P(N|+) = P(+|N)P(N)/P(+) = 0.01*0.9999/0.010098 ~ 0.9901960784313725 P(N|+) ~ 99%
P(S|+) = the probability that a sick person tested positive P(S|+) = P(+|S)P(S)/P(+) = 0.99*0.0001/0.010098 ~ 0.0098039215686275 P(S|+) ~ 1%
P(N|+)/P(S|+) = 101.
You are 101 times more likely to NOT be sick given a positive test result than to BE sick given a positive test result. (Even though the test is 99% sensitive and 99% specific!)
On April 11 2011 14:26 Earll wrote: This might be something everyone knows the answer to and might already have been posted(though I cant say I have seen it and I think i have read all of the 16 pages) But here goes.
You have to walk into 1 of 2 doors, where one will lead to certain instantenous death, and the other will lead to a life full of starcraft 2 and happy things. Each door is guarded by an all knowing all powerfull zergling(A crackling, if you will). One of the zerglings will always lie and the other will always tell the truth. You can ask One of them, 1 question. What do you ask to figure out what door to take? (You obviously do not know what zergling lies, and what door the zergling guards holds no relation to him lying or telling the truth.)
Suppose that 20 starcraft pros played in a tournament where each player competed against every other player exactly once. The result of each game is only win/lose, no draw. Is it always possible (regardless of results) to name the players 1,2,3,...,20 so that player 1 defeated player 2, player 2 defeated player 3,... player 19 defeated player 20? Give a proof or a counterexample.
On April 11 2011 20:41 ]343[ wrote: Hmm, don't think this has been posted yet?
Let A be a finite set of (distinct) integers. Let A+A be the set of all sums a+b where a, b are in A, and similarly define A-A to be the set of all differences a-b where a, b are in A. a and b are not necessarily distinct. Prove or disprove: |A+A| <= |A-A|.
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
1.05 is close, but not the exact time, because as the minute hand move to 1, the hour hand move a little bit forward, and when the minute hand move that bit forward, the hour hand move a tiny bit more, and so on... >:D
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
What? Flip both, flip 11 when 7 finishes? That'd be 7+11. 1. Flip both 2. Let 7 finish. 3. As 7 finishes, flip it. This will leave 4 minutes left on 11, and 7 minutes on 7. 4. 11 finishes, with 3 minutes left on 7. Flip 7. Since it's an hour glass, if there's 3 minutes on one side, there's 4 minutes on the other. 5. When 7 finishes once more, 15 minutes are gone.
A = 5L. B = 3L 1. Fill #1 2. Pour A into B until B is full. Now now we have 2L in A as 5-3 = 2. 3. Empty B and pour the 2L from A into B. We now have the 2L in B instead. Empty A. 4. Fill A completely. (A = 5L), (B = 2L) at this point. 5. Now pour from A until B is filled again. This leaves us with 4L (as 5L minus 1L to fill B = 4L) in A and 3L in B. 6. Empty B and voila! 4L in A.
A = 7min. B = 11min. 1. Flip both. 2. When A runs out of time, flip it again. Remaining time on B is now 4 minutes. (11-7). 3. When B runs out of time, flip A for the third time. (11 min passed. 3/4 -> 3 top, 4 bottom of A.) 4. When A runs out, we got a winner.
1. Fill the 3L jug pour it into the 5L, fill the 3L jug again and fill the 5L jug so you have 1L left in the 3L jug, Empty the 5L and pour the 1L into it, fill the 3L once more and pour into the 5L jug and there you have it, 4 litres of water.
On April 10 2011 00:05 stepover12 wrote: For your curious minds
- Feel free to discuss your thoughts, but please put solution in + Show Spoiler +
here
- Feel free to post your own brainteaser/problem, use bold font so I can find them and paste into original post. - Don't troll plz <3
1. you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
3. You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
5. (my favorite :p) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer.
1. just keep pouring one into the other it'll solve itself eventually. this isn't really a riddle lol lemme try now 5-0 2-3 2-0 0-2 5-2 4-3 yep..
2. number the bags A..J then just weigh A+2B+3C+4D etc
3. flip em both, wait til 7 minute runs out, pop on the eggs, wait til 11 runs out (4 minutes), then re-flip 11 again? this is identical to question #1
4. pretty famous not gonna bother
5. edit: intuitively.. yeah. i don't quite see what the riddle is. let's say it's a 25x1 rectangle with 25 circles side by side. cutting the radius in half would imply that in place of each circle you could fit 4, 4*25=100? what are we looking for here, exactly? what do you mean by 'entirely covered' and why is the 'entirely' absent for the second part of the question?
you have 10 from each bag, if you put 1 from the first bag, 2 from the second bag, 3 from the third etc all on a scale, you can tell how many counterfeit coins by subtracting the total weight from 55 (number of coins), for each counterfeit coin there should be a deposit of 0.1g, the number of coins tells you which bag.
Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes? Also it burns unevenly, so cutting/folding doesn't work. + Show Spoiler +
on 1 string light both ends at the same time light 1 end of the other string, as soon as the first string is completely burned up, light the other end of the second string when the second string is burned up, it's been 45 minutes!
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
What? Flip both, flip 11 when 7 finishes? That'd be 7+11. 1. Flip both 2. Let 7 finish. 3. As 7 finishes, flip it. This will leave 4 minutes left on 11, and 7 minutes on 7. 4. 11 finishes, with 3 minutes left on 7. Flip 7. Since it's an hour glass, if there's 3 minutes on one side, there's 4 minutes on the other. 5. When 7 finishes once more, 15 minutes are gone.
my answer isn't 7+11, it's 4+11 if you read my explanation after the flipping steps, when the 7 finishes for the first time, there is 4 minutes left in the 11 minute hourglass, so if you flip that after it finishes you get 4+11=15
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
What? Flip both, flip 11 when 7 finishes? That'd be 7+11. 1. Flip both 2. Let 7 finish. 3. As 7 finishes, flip it. This will leave 4 minutes left on 11, and 7 minutes on 7. 4. 11 finishes, with 3 minutes left on 7. Flip 7. Since it's an hour glass, if there's 3 minutes on one side, there's 4 minutes on the other. 5. When 7 finishes once more, 15 minutes are gone.
my answer isn't 7+11, it's 4+11 if you read my explanation after the flipping steps, when the 7 finishes for the first time, there is 4 minutes left in the 11 minute hourglass, so if you flip that after it finishes you get 4+11=15
Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes? + Show Spoiler +
on 1 string light both ends at the same time light 1 end of the other string, as soon as the first string is completely burned up, light the other end of the second string when the second string is burned up, it's been 45 minutes!
you're missing the rather important condition that the strings burn unevenly otherwise it's a trivial problem where you can just cut one string into 3/4, 1/4 and burn it without even using the 2nd stirng
about the water one...im guessing that you fill up the 5 liter jug using the 3 liter jug until the 5 liter jug is full and 1 liter of water is left in the three liter jug.
5 liter jug = full 3 liter jug = 1 liter
then you empty the 5 liter jug, pour the 3 liter jug into the 5 liter jug and, after filling the 3 liter jug, pour the 3 liters into the 5 liter jug making 4 amirite?
1. fill the 3 liter and throw those 3 liters in the 5 liter. now fill the 3 liter again and fill the 5 liter. there is now one liter in the 3 liter jug. empty the 5. throw the one liter from the 3 into the 5. fill the 3 liter jug and add to the 5 liter jug. you now have 4 liters.
2. take nothing out of bag 1, take 1 coin out of bag 2, 2 out of bag 3, 3 out of bag 4 etc. put all the bags on the scale. check to see how many weight you miss. if you miss nothing then bag 1 is the counterfeit. if you miss 0,1 gram bag 2 is the counterfeit bag, miss 0,2 bag 3 is fake etc.
3. set the 7 and the 11 minute glass. drop in the egg when the 7 runs out. then flip the 11 minute hourglass when it runs out. egg is done when the 11 is done the second time (4+11)
4. 1:07 but you need some trig to solve this, too lazy to do this. should be pretty easy someone in high school did this in ~3rd year.
5. surface area of the disks: PI*1^2 * 25 = 25PI
other disks: PI* 0,5^2 * 100 = 0,25*100*PI = 25PI . yes you can.
Ahh, for number 5, showing that the area is enough to cover may suggest the answer is yes. But it is not a definite proof that there is a way to cover the rectangle with those smaller disks.
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
What? Flip both, flip 11 when 7 finishes? That'd be 7+11. 1. Flip both 2. Let 7 finish. 3. As 7 finishes, flip it. This will leave 4 minutes left on 11, and 7 minutes on 7. 4. 11 finishes, with 3 minutes left on 7. Flip 7. Since it's an hour glass, if there's 3 minutes on one side, there's 4 minutes on the other. 5. When 7 finishes once more, 15 minutes are gone.
It's not 7+11, you begin boiling then egg when the 7 finishes, leaving 4 minutes on the 11 hour glass, when the four minutes is over, you flip it for another 11 minutes. 4+11 is 15
On April 10 2011 00:05 stepover12 wrote: For your curious minds
- Feel free to discuss your thoughts, but please put solution in + Show Spoiler +
here
- Feel free to post your own brainteaser/problem, use bold font so I can find them and paste into original post. - Don't troll plz <3
1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
fill the 3 liter jug full and pour it in the 5 liter jug then fill the 3 liter jug again and pour till the 5 liter jug is full you now have 1 liter in the 3 liter jug. pour out the 5 liter jug then pour 1 liter of water from the 3 liter jug in to the 5 liter jug and then fill the 3 liter jug one more time and pour it in the 5 liter jug
1: Fill the 5l can, fill the 3 liter with as much as u can from the 5 liter = 2lites left in the 5 liter. empty out the 3l. Pour the 2 liters that are in the 5 liter in to the 3 liter. Fill the five liter and fill the 3 liter with the water from the 5 liter = there will be 4 liters left in the 5 liter can.
2: take 1 coin from the first bag, 2 from the next etc up to the tenth. If none of them were fake they should weigh 55 grams. But since one of the stacks where fake and only weighed 0.9 grams per coin. They will weigh anything between 54.9- 54 and that is how u determine which one is fake since the fake ones will be in the X stack when the weigh is 55- x*0.1
3: start both hourglasses when 7 minute runs out start boiling the egg. That means that it's 4 min left on the 11 minute hour glass. and when it runs out just flip it over again 4+ 11 = 15.
4: 12 hours later?
5: Reducing the radius of an object means that it reduces it's size both in height and width. this being said if if scale this whole question down to it being only 2 disks in the beginning and that the end result would be 4 times more (8) if you reduce the radius to half of it's current state it's real easy to picture it in you're head. half the radius = 3 more disc fit for each existing disc as one is added to it's width and height and one diagonally as this space is freed up as well = there are 3 more per disc than before that there is room for.
6: mind blown :/ gonna think about it...
7: gonna think about it...
8. light up both ends on one of the strings and just one on the other. When the first one burns out light the other end on the second piece of string.
9: 31 - prime 31mod 30 = 1-not prime.
10: yet another one I don't know .p
Here's a real brain teaser though it's a paradox but it was the first paradox that was introduced to me by my math teacher and I thought it was pretty interesting.
"Suppose there is a town with just one male barber; and that every man in the town keeps himself clean-shaven: some by shaving themselves, some by attending the barber. It seems reasonable to imagine that the barber obeys the following rule: He shaves all and only those men in town who do not shave themselves."
For #3, 1) flip both hour glasses over, 2) flip over the 11 minute hourglass when it finishes. The time between when the 7 minute hourglass finishes and the 11 minute hourglass finishes (for the second time) is exactly 15 minutes.
What? Flip both, flip 11 when 7 finishes? That'd be 7+11. 1. Flip both 2. Let 7 finish. 3. As 7 finishes, flip it. This will leave 4 minutes left on 11, and 7 minutes on 7. 4. 11 finishes, with 3 minutes left on 7. Flip 7. Since it's an hour glass, if there's 3 minutes on one side, there's 4 minutes on the other. 5. When 7 finishes once more, 15 minutes are gone.
my answer isn't 7+11, it's 4+11 if you read my explanation after the flipping steps, when the 7 finishes for the first time, there is 4 minutes left in the 11 minute hourglass, so if you flip that after it finishes you get 4+11=15
I got a even more elegant solution. 1)flip both over 2)When 7 min hourglass runs out, flip it over and over again. 3) Let 7 min hourglass finish when 11 min hourglass runs out. + Show Spoiler +
I'm afraid not still. Give any prime who's modulus 30 is 1 and it'll disprove your conjecture. 61 disproves, 151 disproves, 541 disproves, etc. In other words, if (x-1)/30=integer, where x is a prime number, then your idea doesn't work. Maybe exclude 1 and it'll work.
I'm afraid not still. Give any prime who's modulus 30 is 1 and it'll disprove your conjecture. 61 disproves, 151 disproves, 541 disproves, etc. In other words, if (x-1)/30=integer, where x is a prime number, then your idea doesn't work. Maybe exclude 1 and it'll work.
Yes, exactly. Let's say R is not 1, then must R be prime? I was planning an argument in my head and didn't consider the case R=1 hehe, you got me.
On April 10 2011 00:05 stepover12 wrote: 9. (if you know what prime numbers are) If a prime number greater than 30 is divided by 30, you get a remainder R. Must the remainder R be a prime number? Why or why not?
30 = 2*3*5, which means that the remainder can't have any of those factors, else: Let p be a prime > 30 p mod 30 = R => p = 30*n + R if R = m*5 p = 30x + 5m = 5*(6x+m) => 5|p => p can't be a prime if it has 5 as a factor (we assume p>30) same for R=2m and R=3m So the only possible remainders are: 1, 7, 11, 13, 17, 19, 23, 29 1 happens to be not a prime. 91 mod 30 = 1 ;o
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
1. fill 5, pour 5 into 3, empty 3, pour 5 into 3, fill 5, pour 5 into 3. 5 jug now has 4 in it.
3. flip 7 and 11. when 7 is done, flip 7 again. when 11 is done, flip 7 (with 3 on it) and 11. when 3 is done, neutralize 11 (8 on it.) get egg in pan, flip the 8, when 8 is done, flip 7
4. if both hands are infinitely thin then they will always coincide but the real answer you seek is some point near 1:05 that i am too lazy to calculate. you just have to remember it isn't 1:05 because the hour hand will have moved up 1/12 of the distance between 1 and 2 at that point.
7. yes. weigh 3 on one side and 3 on the other. if one side weighs less, that advances to step two. if both sides are equal, the unweighed 3 coins advance to step 2. then in step 2, weigh any 2 random coins. if equal, its the unweighted 1. otherwise its on the scale.
Weigh 3 coins on each side. If they weight the same, then none were counterfeit and you have 3 coins left. You would then choose 2 of those coins and weigh them. If they're the same, then the third one is counterfeit. If not, then it's the lighter one.
If the initial weighing gives you that one side is lighter, then you have 3 coins, one of which is counterfeit. You can just choose 2 and weigh them. If they weight the same, then the third one is counterfeit. If not, then it's the lighter one.
The first person calls out red if he sees an odd number of red hats and blue if he sees an even number. This person might die. But he calls out red, the next person looks back and if he sees an odd number, then he must be wearing blue. If not, then he must be wearing red.
This is a simple inductive reasoning problem. This means we have to take it one step at a time, starting with 1 blue-eyed person.
1. If there is 1 person he leaves on the first night because he knows he must be the only one with blue eyes.
2. If there are 2 people, each one of them sees each other. The next day, they wonder why the other person didn't leave, because if they were the only blue-eyed person, they would have left. Both of them being perfectly logical come to this conclusion and leave on the second night.
3. If there are 3 people, each one of them sees each other. The second day, no one left, because if there were 2 people they would have to wait until the second night anyways. The third day, they realize that none of them left because there must be 3 people on the island with blue eyes, because otherwise if there were 2 people then the 2 would have left on the second night. So, they leave on the third night.
... So on and so forth until...
100. If there are 100 people, all of them see 99 people. The 100th day, no one left, because if there were only 99 people on the island, they would have left on the 99th night. So, the 100 people leave on the 100th night.
Just to add for Day9's strings aka problem 8. The standard solution provided still has one nitpicky assumption that speed of burning is independent of the direction of burning. I did not think about it too much, but I suspect that without that assumption it has no solution, but if someone has any, I would love to hear it, I will try to prove that there is none in the meantime
Well the question states "in some way",so i guess we have to follow the assumption that it can somehow fill up the whole rectangle without overlaps.A somewhat unrealistic question if u ask me.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
Obviously it will occur after 1:00. The minute hand will be slightly past 5 as someone mentioned earlier in thread. This is just a basic system of equations math problem.
Variables: HA = hand angle, MA = Minute Angle, m = number of minutes past 1
HA = 30 + (30 m / 60), where m is the number of minutes that have passed. MA = 6 m.
Condition: HA = MA HA = 30 + MA /12 => MA = 30 + MA / 12 Solving: 11 MA = 360 => MA = 360 / 11
m = MA / 6 => m = 60 / 11
Therefore, they overlap in 1 hour and 60/11 minutes (approx 5 min 27 sec)
5. (my favorite ) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer.
Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap.
On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
Cool Question.
Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them.
The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition.
If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see.
If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color.
The viziers will agree on all this on the night before the hat day.
More elegant solution (same idea): see Murderotica's post a couple posts down.
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
On April 10 2011 01:27 mcc wrote: Just to add for Day9's strings aka problem 8. The standard solution provided still has one nitpicky assumption that speed of burning is independent of the direction of burning. I did not think about it too much, but I suspect that without that assumption it has no solution, but if someone has any, I would love to hear it, I will try to prove that there is none in the meantime
There is a solution for the string to only burn in one direction:
3. Fold that string in half again, folding over the top.
4. This will cause there to be 2 quarters of string side by side and 2 on top of those 2. Unfold the string from the quarter that is adjacent to the second string, and put the unfolded string perpendicularly over the other string.
5. Light the extended string. When the flame ignites the second string which is on top of the first (perpendicularly), the allotted time has passed.
Well the question states "in some way",so i guess we have to follow the assumption that it can somehow fill up the whole rectangle without overlaps.A somewhat unrealistic question if u ask me.
The original covering by 25 disks must have overlaps. And the covering by100 disks (is it exists) should have overlaps as well.
Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap.
On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
Cool Question.
Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them.
The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition.
If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see.
If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color.
The viziers will agree on all this on the night before the hat day.
That solution can be simplified. We do not need to know whether the amount of viziers is even or odd. They can assert that the number of blue hats will be odd, whether it is or not. So, if the first vizier sees an odd number of hats, he will say he has a red hat to keep the number odd. The next vizier will know that the guy behind him said that his hat was red because he sees an odd number of blue hats ahead of him, and from this information he can deduce what hat is on his head after viewing if the number of hats ahead of him is odd or even. This is more elegant but you were of course still correct.
I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
For example, if we had 2 blue (A & B) 2 brown (C & D) 1 green (guru):
Everyone knows: -at least 1 blue (Guru statement is key to this problem) -Therefore, we can assume that if there is a person who sees no others with blue eyes, he will leave knowing he has blue eyes.
NIGHT 1: Nothing happens A & B: See 1 blue 2 brown 1 green. Know that there are at least 1 blue. No conclusions can be made.
C & D: See 2 blue 1 brown 1 green. Know that there are at least 1 blue. No conclusion can be made.
Guru:: 2 blue 2 brown. No conclusion.
NIGHT 2: A & B leave. A & B: See 1 blue 2 brown 1 green. Noone left on the 1st night, therefore they can imply there are >1 persons with blue eyes & therefore they must have blue eyes. Both of them leave.
C & D: See 2 blue 1 brown 1 green. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes.
Guru: 2 brown 2 blue. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes.
NIGHT 3: C & D 1 green 1 brown. C & D can conclude that their eyes were not blue since A & B have left together. This doesn't help since both C & D cannot deduce if there are 2 or 1 brown eyed people remaining. For all they know their own eyes are red and as such the information from the 1st and 2nd nights is useless.
Guru: The guru cannot get any further useful information either since there is no information pertaining to green eyes available to him.
Given this example we can begin to see that if there are X blue eyed people they will leave the island on night X, but only if they know there is at least one blue eyed person. We can then conclude that all of the blue eyed people will leave on the 100th night.
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
Uh well... These switches are outside the room right?
Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap.
On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
Cool Question.
Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them.
The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition.
If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see.
If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color.
The viziers will agree on all this on the night before the hat day.
That solution can be simplified. We do not need to know whether the amount of viziers is even or odd. They can assert that the number of blue hats will be odd, whether it is or not. So, if the first vizier sees an odd number of hats, he will say he has a red hat to keep the number odd. The next vizier will know that the guy behind him said that his hat was red because he sees an odd number of blue hats ahead of him, and from this information he can deduce what hat is on his head after viewing if the number of hats ahead of him is odd or even. This is more elegant but you were of course still correct.
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
I have seen this before and I think it is a terrible question. It is not a math/logic type brainteaser.
The solution involves keeping one light on for a while and then turning it off so it remains hot when you go in, so you can feel it (and turn on another one before going in). Kinda lame imo
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
Turn the first switch on and wait for a few minutes, then turn it off again. Turn the second switch on and enter the room. If the light is on you know it's switch 2 If the light is off but the bulb is hot it's switch 1 If the light is off and the bulb is cold it's switch 3
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free. let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him. On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how?
REMEMBER/rules 1) there are 2 black balls in the bag and they are not changed 2) he does not have a trick ball anywhere - he chooses a ball from the bag 3) the ball he chooses is the one he is represented by 4)where he is
Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this): 1. + Show Spoiler +
Fill the 5ltr jug, use it to fill the 3ltr jug. You now have 2ltr left in the 5 ltr jug. Empty the 3ltr jug, pour the 2ltr from the 5 ltr jug into it. The 3ltr jug now has place for 1ltr. Fill the 5 ltr jug, and then, with it fill the 3ltr jug until it is full. You now have 4ltrs in the 5 ltr jug.
It is possible. You take one coin from the first, 2 coins from the second, three coins from the third, etc until 10 coins from the 10th. If you weigh the coins you took, you will get a weight of 55gr (55 coins), -0.1gr for each counterfit coin. If you have one counterfit coin, it will weigh 54.9, for 2 it'll be 54.8, etc until 10 will be 54gr. According to the number of counterfit coins, you will know from which bag you took them.
Flip the 7 and 11 minute hourglasses simultaneously. Start cooking once the 7min one runs out. That leaves 4 minutes until the 11 minute hourglass is empty. Once it empties 4 minutes later, flip it over. Once it is done, you have been cooking for exactly 4 + 11 = 15 minutes.
Let's assume that they move at an exact uniform speed (won't skip eachother). In the next 12 hours, they will meet 11 times (once at 1pm and a bit, once at 2pm and a little more, etc, until the meet again exactly at 12am). Therefor, in 12 hours they meet 11 times. Since they move at uniform speeds, they must meet once every 12/11 hours, meaning 1 hour, 5 minutes, 27 seconds, and 3/11ths of a second (if my arithmetic is correct).
It can be found in 2 comparisons like this: Split into 3 piles of 3, compare 2 of them on the scale. If one is lighter, then in contains the coin. Otherwise, it is in the 3rd pile. Hence, the coin is in the lightest pile. Take 2 coins from that pile, compare them on the scale, if one is lighter, that is the coin you are looking for, otherwise, it is the coin that wasn't on the scale.
You take the first string and light it on both sides so it will burn twice as fast. You light the other string on one side at the same time. Once the string lighted on both sides finishes burning, exactly 30 minutes have passed, so start burning the other string on the other side, so it will burn twice as fast. Since the remaining 30 minutes of burn times have been halved, it will finish burning in another 15 minutes. Once it finishes burning, 45 minutes have passed.
Let n be a prime number over 30. Let m be the remainder of dividing n by 30, and p is the result rounded down. m is therefor a non-prime number under 30.
Let's assume that m is not a prime number. Therefor, n = 30*p + m. Hence, p = (n-m)/30. If m is even, therefor, n has to be even, since an odd number minus an even number is an odd number, and p is by definition a whole number. If m is divisible by 3, then so is n. If m is divisible by 5, so is n, for the same reason. Hence, m has to be a multiplication of prime numbers greater than 5. Hence, m is equal to or greater than 7*7 = 49, which is impossible as it is the remainder for division by 30.
Hence, the remainder has to be a prime number.
5, 6 and 10 will take time I don't have if I want to get stuff done before TSL, so it will have to wait.
Sure they can. The rectangle can be filled with 25, so a rectangle with half the height and half the width can be filled with circles with half the radius in the exact same way. Therefor, you can put 25 of each of the smaller discs in each quarter of the rectangle using the exact same strategy to fill it. Feel stupid that I didn't get that instantly :/
Edit 2: OK. The first 1-4 and 7-9 were all instant, and came to me after typing the rest. So that's 21 minutes. ARG!!@# 3 more now!
The first vizier says his hat is red if he counts an odd number of blues on the others. Therefor, if he is correct, everyone knows there is an odd number of blues and can deduce that if he sees an even number of blues his hat is blue, and otherwise it is red. If he is incorrect, therefor there is an even number of blue hats, and the other viziers can (in similar fashion) deduce the color of their hat.
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
The first line can't cross any. The second line can at most cross the first. The third can cross the other two at most. And so on. Hence: the maximum number of crossings is 0+1+2+3+...+9, therefor, 45 possible intersections.
Hooray for another 10 minutes spent! 6 and 11 will take some time, and I want to get food ready for TSL :D
Start both, flip the 7m when it runs out after 7m, flip it again in 4 minutes when the 11m runs out, it will then run out 4 minutes later for a total of 15 minutes.
100 disks of radius 1 can cover 4 same sized rectangles, shrinking everything to .5 scale you have 1 rectangle of the original size covered by 100 disks of radius .5
On the first night if someone saw noone else with blue eyes they would know they had the blue eyes and they could leave. But since that doesn't happen everyone knows that noone sees 'no one with blue eyes'; ie. everyone sees at least 1 person with blue eyes. So on the second night someone who saw only one person with blue eyes would know that they also had blue eyes... So one the 100th night everyone who sees 99 people with blue eyes will know that they have blue eyes, so all 100 of them leave that night.
Weigh 3 against 3, if one weighs less it contains the counterfeit. If it balances then the counterfeit is in the 3 not weighed. Weigh 1 vs 1 of the last 3.
Light three of the ends at the same time. The one with 2 lit ends should burn in 30 minutes. Then light the other end of the remaining string and it should burn out in 15 minutes.
If the remainder was a multiple of 2,3,5 then it wouldn't have been prime before dividing by 30. So the only remainders possible when starting with a prime number are 7, 11, 13, 17, 19, 23, 29 which are all prime.
If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members.
12 coins, 1 of which is either heavy or light EEEEEEEEEEEE (24 possibilities), measure 4v4 they balance, the 4 others are either heavy or light ___EEEE (8 possibilities), measure 3 versus 3 legit ___ they balance, the one you didn't measure is either heavy or light ______E (2 possibilities), measure it versus 1 legit ___ they don't, you have 3 heavy or 3 light ______hhh or lll (3 possibilities), compare 2 ______ they balance, the third is the fake ______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light ___hhhhllll (8 possibilities), measure hhl versus hhl ___ they balance, one of the 2 lights you didn't measure is fake ______ll (2 possibilities), measure versus each other ___ they don't, one of the 2 heavies was heavy or the opposite light was light ______hhl (3 possibilities), measure the heavies versus each other ______ they balance, the light was light ______ they don't, the heavy one is heavy
i did have to proof #7 back in grade 9 (kind high level math genius back then :D)
its not that simple to solve it since you will have to alternate the coins in the set you used to weight the 1st time.
The REAL #7 should be that you DONT know if the fake coin weight MORE or LESS than the real coins. It makes the problem WAYYYYY more complicated but still solve-able (hoping the OP gona put this one up :D)
There are 12 coins. One of them is fake: it is either lighter or heavier than a normal coin. Find the fake coin and tell if it is lighter or heavier by using a balance the minimum number of times possible. (answer is 3)
we know immediately that the hands will touch again sometime after one. so i set up two equations, y=x which is the movement of the minute hand at one, and y=x/12 + 5 which is the hour hand at one. from there i just found out where the two lines intersect, which happens 5.45 minutes after the clock hits one.
2.b. Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though.
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice but don't open it straight away and the presenter opens one of the other 2 doors you didnt choose to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
For the OP and others, yes this is the monty hall problem.
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free. let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him. On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how?
REMEMBER/rules 1) there are 2 black balls in the bag and they are not changed 2) he does not have a trick ball anywhere - he chooses a ball from the bag 3) the ball he chooses is the one he is represented by 4)where he is
for #6 everyone so far is WRONG (or at least, their answer is incomplete). + Show Spoiler +
So if 100 blue eyed people leave on the 100th night, on the 101th day, all the brown eyed people would've realised that the only reason the 100 blue eyed people would've left is because they themselves (the blue eyed people) had realised that there are ONLY 100 blue eyed people on the island, and therefore had to leave. After this, the brown eyed people would all realise that each of themselves is also brown-eyed and would all leave on the 101st night.
On April 10 2011 01:53 funk100 wrote: I have one + Show Spoiler +
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free. let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him. On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how?
REMEMBER/rules 1) there are 2 black balls in the bag and they are not changed 2) he does not have a trick ball anywhere - he chooses a ball from the bag 3) the ball he chooses is the one he is represented by 4)where he is
He picks a ball, drops it off the cliff 'by accident' and says 'oops'. But to find the colour of the ball he chose, they look for the other ball that is remaining in the bag. A black ball is remaining so they all agree he chose the white ball.
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. e You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes. If you switch you have a 66.6% chance of winning the car, and if you don't switch you have a 33.3% chance. In the beginning when you picked the first door, you had a 66.6% chance of pickng a goat, and a 33.3% chance of picking the car. If you picked a goat (66.6%), he opens a goat door, so the other door has the car. If you chose a car first (33.3%), he opens a door with a goat, and the other door has a goat.
5. Yes. A circle of radius 1 covers a square of exactly 2, since the width of the square is square root of 2, found using pythagorean theorem. A circle of radius 1/2, covers a square of exactly 1/2, using same calculations.
So 25 * 2 = 100 * 1/2 = 50. Both sets of circles can cover a rectangle of area up to 50.
fill 5L jug, pour until full into 3L jug, empty 3L jug, pour the rest of the 5L jug into the 3L jug, fill 5L jug again, poor 1L to fill the 3L jug, than you have 4L
On April 10 2011 02:03 GQz wrote: for #6 everyone so far is WRONG (or at least, their answer is incomplete). + Show Spoiler +
So if 100 blue eyed people leave on the 100th night, on the 101th day, all the brown eyed people would've realised that the only reason the 100 blue eyed people would've left is because they themselves (the blue eyed people) had realised that there are ONLY 100 blue eyed people on the island, and therefore had to leave. After this, the brown eyed people would all realise that each of themselves is also brown-eyed and would all leave on the 101st night.
no, they could have red eyes (or any other eye colour) they don't know there are only brown and blue eyed persons.
1) This was an easy one. Fill up the 3 liter bottle. Pour into the 5 liter bottle. Fill the 3 liter bottle again. Fill the 5 liter bottle, you will have 1 liter left over. Empty the 5 liter bottle. Pour the 1 liter left over into the 5 liter bottle. Fill up the 3 liter bottle. Pour the 3 liters into the 5 liter bottle. Exactly 4 liters.
On April 10 2011 02:03 GQz wrote: for #6 everyone so far is WRONG (or at least, their answer is incomplete). + Show Spoiler +
So if 100 blue eyed people leave on the 100th night, on the 101th day, all the brown eyed people would've realised that the only reason the 100 blue eyed people would've left is because they themselves (the blue eyed people) had realised that there are ONLY 100 blue eyed people on the island, and therefore had to leave. After this, the brown eyed people would all realise that each of themselves is also brown-eyed and would all leave on the 101st night.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The ONLY thing YOU know is that SOMEONE on the island has blue eyes, nothing else
On the first night if someone saw noone else with blue eyes they would know they had the blue eyes and they could leave. But since that doesn't happen everyone knows that noone sees 'no one with blue eyes'; ie. everyone sees at least 1 person with blue eyes. So on the second night someone who saw only one person with blue eyes would know that they also had blue eyes... So on the 100th night everyone who sees 99 people with blue eyes will know that they have blue eyes, so all 100 of them leave that night.
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. e You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes. If you switch you have a 66.6% chance of winning the car, and if you don't switch you have a 33.3% chance. In the beginning when you picked the first door, you had a 66.6% chance of pickng a goat, and a 33.3% chance of picking the car. If you picked a goat (66.6%), he opens a goat door, so the other door has the car. If you chose a car first (33.3%), he opens a door with a goat, and the other door has a goat.
To be horribly pedantic, since I'm bored, the question ought to state that the presenter always gives the option to switch doors.
Perhaps he only gives you the option to switch when you've chosen the correct door? In that case you should never switch. Or perhaps he gives the option to switch 2/3 of the time when you choose the correct door, and 1/3 of the time when you choose the incorrect door? In this case you would be indifferent to switching, having a 50% chance of being right whether you switch or not.
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. e You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes. If you switch you have a 66.6% chance of winning the car, and if you don't switch you have a 33.3% chance. In the beginning when you picked the first door, you had a 66.6% chance of pickng a goat, and a 33.3% chance of picking the car. If you picked a goat (66.6%), he opens a goat door, so the other door has the car. If you chose a car first (33.3%), he opens a door with a goat, and the other door has a goat.
To be horribly pedantic, since I'm bored, the question ought to state that the presenter always gives the option to switch doors.
Perhaps he only gives you the option to switch when you've chosen the correct door? In that case you should never switch. Or perhaps he gives the option to switch 2/3 of the time when you choose the correct door, and 1/3 of the time when you choose the incorrect door? In this case you would be indifferent to switching, having a 50% chance of being right whether you switch or not.
Well, I do suppose I'm assuming he asks you if you want to switch every time.
On April 10 2011 02:45 susySquark wrote: 5 is either poorly worded or too vague. How big's the rectangle??
I.E. If the rectangle is the size 2 inches by 2 inches, 25 1 inch circles can easily cover it, and 25 1/2 inch circles can also obviously cover it.
The size of this rectangle is unknown, we only know that it can be covered by 25 disks or radius 1. This vagueness adds to the elegance of the solution, i.e. we don't know much about the rectangle, but we can still say something meaningful about it.
OK, let's take a strict mathematical solution here: a) If there had been one person with blue eyes, that person would only see people with brown eyes and leave on the first night.
b) Let's assume that if there were n people with blue eyes, they'd all leave on the nth night. If there were n+1 people, then each person with blue eyes would see n others with blue eyes. Since they hadn't left after n nights, therefor, they know for a fact there are n+1 people with blue eyes, and that they are one of them.
By induction, for every number of people with blue eyes, they would all leave on the night numbering the exact same amount of people since the Guru spoke.
The rest could not leave, because as far as they know, they have green or purple or whatever colored eyes. There would not be enough information for any of the rest to leave.
11. I'll probably think up an induction or proof by contradiction for it soon.
I will do my best to explain this with only text, as drawings are generally necessary to convey the logical processes. It may be easier to use Proof by induction, which is also possible (I have tried both ways)
For a circles (of similar size) to actually entirely 'cover' a continuous rectangle, we note that it must intersect (overlap) with the other circles. At the extremes, it must intersect exactly at the edge of the rectangle to ensure an airtight cover. So from this we have our first postulate:
(1) All circles must intersect other circles at the edge of the rectangle (for maximum area).
Secondly, we note that rectangles can be split up into other, smaller rectangles. In fact, we can split them up such that they all become rectangles within the circles (that is, rectangles touching the edges of the circles). These squares will exist with their vertices at the intersections of the circles.
2) Rectangles can be described as a 'mosaic' of smaller rectangles.
3) Rectangles that are fully covered by circles have vertices that rest on (or inside) a circle's edges.
So from these postulates, we can bring it down to whether 4 circles of radius 0.5 can cover any rectangle that fits within a circle with a radius of 1. The result of this would be repeated as the rectangles (and overlapping circles) could then be tessellated to form some sort of rectangle. For this, we need only consider the 2 extreme cases - the square (one of the extremes) and the flat line (the other). If we draw a cartesian plane with a unit circle with center at the origin, we can see that the square is defeated by placement of the smaller circles with centers (.5,.5),(.5,-.5), (-.5,.5), (-.5,-.5). The flat line should be obvious and needs no explaining.
So from 1), 2) and 3), we deduce that if 4 small circles can cover any rectangle inside one big circle, then we can repeat the logic for any larger rectangle comprised of a multiple of these smaller rectangles. And from above, we have shown that any rectangle in a larger circle can be covered by 4 smaller circles. So we can deduce that any rectangle made out of 25 rectangles that can each be covered by a unit circle, it can also be covered by 100 circles with radius 0.5.
PROOF BY INDUCTION (An easier explanation) Required to prove xL<=4xS (where <= denotes that 4xS can encompass that same Area or more, than xL in relation to a rectangle as. Works the same as 'greater than or equal to' in mathematics)
Step1. It is true for x=1 L<=4S (where L is a Large circle, 4S is 4 smaller circles) That is, what we have proven already, that a rectangle that fits in a unit circle L can be covered by 4 smaller circles S.
Step 2. Assuming it is true for x=k, prove true for x=k+1 Because 4S>=L (step 1) and 4kS>=kL (Assumption) 4S+4kS>=L+kL (Addition of equations) therefore, 4(k+1)S>=4(k+1)L
Step 3. Therefore, because it is true for x=1, it is also true for x=1+1=2+1=3+1=4 and so on.
Therefore, it is true for x=25. 100S>=25L Any rectangle encompassed by 25 Larger circles can be encompassed by 100 smaller circles (QED)
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
I dislike this problem. The classical solution only works if the lightbulb is inefficient enough to generate more heat and subsequently store it in the material of the bulb such that an amount that is above the human threshold for detection remains after the convection that occurs between the time when you switch it off and walk over to it. LED's lol over this method, which might make this riddle obsolete in a few years
The answer is yes because any prime number divided by a relative prime gives a remainder of another prime number
The remainder of 109 divided by 60 is 49, which isn't prime.
It's because no odd composite number between 2 and 30 fails to have a common denominator with 30. It is impossible to form a composite number with a remainder of 9, 15, 21, 25, or 27 after dividing it by 30.
It's also difficult to understand your latest problem:
15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
Which 100 distinct integers? It's theoretically possible to have a collection of 100 distinct integers in which this kind of selection is impossible (100 multiples of 13, for instance. Only multiples of 91 are then selectable.)
7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
1) Light stick 1 in one end. 2) Light stick 2 in both ends. 3) When stick 2 has finished burning, it has passed 30 minutes. So stick 1 will burn for 30 minutes more. 4) Light stick 1 in the other end = 15 minutes
People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?
Which 100 distinct integers? It's theoretically possible to have a collection of 100 distinct integers in which this kind of selection is impossible (100 multiples of 13, for instance. Only multiples of 91 are then selectable.)
If you had the first 99 multiples of 13, then 15 of them would have differences of 91. (14*7+1=99) Not sure what language to use to go about proving that though.
On April 10 2011 04:04 Manit0u wrote: People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?
#1: Fill up the 5 liter jar completely. Pour from 5l to 3l until 3l is full. You now have 2l left in the 5l jug.
Empty 3l jug. Put remaining 2l in the 3l jug. Fill up 5l jug. You now have 2 liters in the 3l jug, and 5 liters in the 5l jug. Now just fill water from the 5l into the 3l jug until the 3l jug is full. You now have 4l in the 5l jug.
#2 Yes. If it is a perfect scale, it is precise enough to determine the difference in the gravitational force that acts between the earth and the bags. I'm not going to give you the exact equations now, but I can describe it: You stack the bags ontop of each other, and weigh them. Each configuration is going to correspond to one number showing up on the scale (the highest being the one when the 0.9x is on the top of the stack, the lowest one being when the 0.9x bag is at the bottom of the stack), which you can use to determine where the bag is in the stack.
#2b Yes. You again stack them up and weigh them all together. Using the same principle as above, the several possible combinations will correspond to a certain set of numbers which show up on the scale. gg
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it:
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room?
Turn on switch #1 on and leave it so for some time - say 10 minutes. Turn it off and turn on switch #2.
Go in the other room. If the bulb is on, is connected to switch #2. If it is off, but it's cold, then it's connected to switch #3. If it is off, but it's hot (when touched), then that's it's indicator, the bulb has been on a bit ago, hence it is connected to switch #1.
Standard "think outside of the box". I want a teddy bear now...at least virtual one.
If you have 7 * (1 - 1) + 2 distinct numbers, you can select 1 so that any choice of 2 selected numbers will have a difference divisible by 7.
Let's assume that if you have 7 * (n - 1) +2 distinct integers which if you select n of them, the difference of any 2 of those n can be divided by 7.
If you have 7 * n + 2 such integers, you can choose 7 * (n-1) + 2 of them so that out of the remaining 7, you cannot select 2 in which their difference will be divisible by 7 (this is possible, because if it wasn't, you could choose 2 or more out of them, and then would be able to choose less from the rest with ease). Therefor, you can select n from the 7*(n-1) + 2 of them, and you need to select one of the remaining 7. As they are of the forms 7a, 7b+1, 7c+2, 7d+3, 7e+4, 7f+5, 7g+6, at least one of them will be selectable so that the difference between it and the rest will be divisible by 7.
By induction, for any natural number n, if you have a group of 7*(n-1)+2 distinct integers, you can select n of them so that the difference between any two will be divisible by 7.
And specifically, for a set of 7 * (15 - 1) + 2 (which equals 100) distinct integers, you can select 15 of them so that the difference between any two will be divisible by 7.
You take one from the first bag, 2 from the second, 4 from the third and 8 from the fourth. Therefor, the weight will be 15 - 0.1n grams, n being the number of fake coins. switch n to a binary representation. It will be a 4 digit binary number, and from left to right, if there is a one in that place, the 4th, 3rd, 2nd, or 1st bag will be of fake coins.
If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members.
Light both ends of one rope and one end of the other. The rope lit at both ends will take half an hour to burn down, when this happens light the other end of the remaining rope which will now take fifteen minutes to burn out. Hey Presto, 45 minutes =D
2 lines gives 1 intersections 3 lines gives 3 intersections 4 lines gives 6 intersections For n lines, the number of intersections is equal to the n-1th triangle number. So for 10 lines, you can have 9(10)/2 or 45 intersections.
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
Interaction is not allowed in any way or form. There's a real solution + Show Spoiler +
It starts with the assumption that if only 1 person has blue eyes, he will leave on the first night because he doesn't see anybody else with blue eyes. You can work your way up from there on. If there are 2 people with blue eyes, they will both see 1 guy with blue eyes. Assuming that the 1 guy with blue eyes would leave if he wouldn't see any other guy with blue eyes around, it has to be assumed that both of them have blue eyes, making them leave on day 2. Note that in the case of 2 blue eyes people, every brown eyed guy will see 2 blue eyed people and the rest brown eyed while the blue eyed guys will see 1 blue eyed and the rest brown eyed.
I'm sorry, my english isn't as good and I'm a little drunk^_^..
If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members.
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
Interaction is not allowed in any way or form. There's a real solution + Show Spoiler +
It starts with the assumption that if only 1 person has blue eyes, he will leave on the first night because he doesn't see anybody else with blue eyes. You can work your way up from there on. If there are 2 people with blue eyes, they will both see 1 guy with blue eyes. Assuming that the 1 guy with blue eyes would leave if he wouldn't see any other guy with blue eyes around, it has to be assumed that both of them have blue eyes, making them leave on day 2. Note that all the other guy will see only brown-eyed people.
I'm sorry, my english isn't as good and I'm a little drunk^_^..
Yeah, that works. So all the blue eyes will leave on n+1 night, all the browns (plus the guru) on n+2 night, where n is the number of people with blue eyes.
But I wanted a better solution, so i read it quite carefully. It does NOT say that interaction in way or form is forbidden. It says the communication is not allowed, hence my version is allowed and everybody can figure out the color of their eyes in one night.
Also, the last person's eye's color can be determined by some switching around and does not obligatorily lies on counting numbers.
On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
are these being added to the OP? I think that would be a good idea.
People are seated around the round table in regular intervals (spaced equally to fill entire table). If 9th person is sitting directly across 22nd, how many people are seated at the table?
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
[QUOTE]On April 10 2011 05:08 shadowy wrote: [QUOTE]On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
[QUOTE]On April 10 2011 05:23 Kazius wrote: [QUOTE]On April 10 2011 05:08 shadowy wrote: [QUOTE]On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3. Once one of the doors is open a goat is revealed, there is a new cituation at hand. Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
[QUOTE]On April 10 2011 05:27 shadowy wrote: [QUOTE]On April 10 2011 05:23 Kazius wrote: [QUOTE]On April 10 2011 05:08 shadowy wrote: [QUOTE]On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3. Once one of the doors is open a goat is revealed, there is a new cituation at hand. Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
On April 10 2011 05:16 cmpcmp wrote: You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
are these being added to the OP? I think that would be a good idea.
I give up. I don't see how to do it with only one question to only one of the sisters and not stated, but provided that I can tell the age of the princess by visually looking at them. I will be very glad if post ( or PM) with a solution.
there is a 2/3 chance of getting the car if you switch (if you don't believe me read the explanation in the next spoiler, or try to figure it out before you do that if you like)
Step 1: Put most simply: your odds of having chosen a wrong door on the first guess is clearly 2/3 because there are three doors and you have no useful information about any of them. Step 2: When the host opens another door for you, assuming you chose wrong on the first guess, if you switch you will end up with the car because it will be the other door.
Essentially, because you know that your odds of picking wrong on the first guess were 2/3 you have information that helps you with your second guess.
On April 10 2011 05:16 cmpcmp wrote: You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
are these being added to the OP? I think that would be a good idea.
I give up. I don't see how to do it with only one question to only one of the sisters and not stated, but provided that I can tell the age of the princess by visually looking at them. I will be very glad if post ( or PM) with a solution.
i dont see any possible solution here, because the middle sister could in a very unlucky coincidence completly behave like one of her other sisters. so its logically impossible to seperate them
[QUOTE]On April 10 2011 05:30 Talanthalos wrote: [QUOTE]On April 10 2011 05:27 shadowy wrote: [QUOTE]On April 10 2011 05:23 Kazius wrote: [QUOTE]On April 10 2011 05:08 shadowy wrote: [QUOTE]On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
In the begining there were 3 doors. So the chance to get the car was 1/3 and a goat 2/3. Once one of the doors is open a goat is revealed, there is a new cituation at hand. Although, not by magic, but by the actions of the person opening that door, now you have one-in-two chance to get a car. 1/3 or 2/3 chance can not exist anymore, as there are no more 3 choices. You get one try, from two choices, hence 1/2 chance or 50/50.
A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Its a while since i heard this one, but im pretty sure the wording is correct..
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
On April 10 2011 05:16 cmpcmp wrote: You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
are these being added to the OP? I think that would be a good idea.
Because of the random elements the middle sister has it is not possible to find out in 1 question which sister is not her (that's all that really matters).
Are you the oldest sister? yes, ?, yes
Are you the youngest sister? No, ?, No
Would you marry the middle sister? No, ?, Yes
Because their is no reasoning behind the middle sisters lies/honesty it is impossible.
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
Ah ok.... I was under the assumption that you are asking all 3 daughters at once and thus this line of questioning was not possible. Woops
You ask any of the daughters "which of your 2 sisters is the youngest?" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
I am very sorry, but this solution does not corresponds to the question being only YES or NO.
you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem."
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
The whole grouping thing serves no purpose whatsoever. If they knew the totals(and it clearly says that they don't) then it wouldn't be a puzzle at all, since the difference between the colors of everyone else and the totals is obviously their own eye color.
Sorry my bad, but that doesn't change the answer in a meaningful way because that question can be asked in a "yes or no" manner. I edited the answer to reflect this.
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
"Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph."
Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks.
you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem."
Yeah, the puzzle becomes obvious when you can ask the middle daughter something about the other daughters. Once you think of a question that never points to the middle daughter when asking the other daughters you win!
A room has 100 boxes labeled 1 thru 100. The names of 100 prisoners have been placed in these boxes by the warden. The prisoners shall visit the room one by one. Each prisoner is allowed to inspect the contents of at most 50 boxes, one after the other and leave the room with no communication with other prisoners. If the prisoner discovers his own name in the boxes he inspects, he is released. The prisoners are allowed to collude before hand and devise a strategy to maximize the chances of releasing each and every prisoner. What is their strategy?
I (think) understand everybody's logic... and it makes sense if you follow those strict mathematical theorms.
What I don't understand is why they are concluding they have blue eyes when they can see other people with blue eyes. If everybody with blue eyes can see someone with blue eyes, nobody will ever leave the island.
I don't see how after 100 days that suddenly means they have blue eyes.
If the ferry takes somebody away every night... and on the first night after the guru speaks nobody leaves... does that not break the condition of somebody had to have left? Which means for 99 days you are counting 99 (or 100) blue eyes on the island. On the 100th day you realize, wow nobody is leaving.... I must have blue eyes? But wait, theres still 99 people with blue eyes here why aren't they leaving? What is so special about 100?
This is probably all jumbled up so don't be too hard on me... I can see why it works if you follow those definitions from the solution of the website.... But only if you assume those theorems are true. They have not given "proof" mind you. Something mathematical textbooks ALWAYS put in a coloured box after a theorem like that.
EDIT: reread the part about people leaving; they don't have to leave....
Here's my train of thought... (I have blue eyes)
Me:"Hmm, 201 people (200 - guru), 100 brown eyes, 99 blue eyes, hmm, I wonder what colour mine are?"
Guru:"SOMEBODY HAS BLUE EYES"
Me: "Yup, this is true, what colour are mine?"
I don't believe this ground up induction works considering you don't start with 1 person. Therefore cannot assume that on the Nth night N people will leave.
AHHHH wtf man stupid ass problem... I see why everyone understands it but it seems like theres is assumptions made by the logicians.
1 person /w blue --> understandable 2 people /w blue---> understandable >3 people /w blue ---> ????
The 1 person example; no other blue eyes, obviously. The 2 people example; the 2nd person knows the first person can see a 2nd set of blue eyes. easy peazy. The 3 people example; the 3rd person knows the other two can see each others eyes. Why would anybody leave?
you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem."
Yeah, the puzzle becomes obvious when you can ask the middle daughter something about the other daughters. Once you think of a question that never points to the middle daughter when asking the other daughters you win!
I am sorry, I still don't see, how there could be a question, that will evade the middle sister problem with a YES/NO question, since her answer will always be random, hence there can no be solution. I personally, can not think about
@cmpcmp Yeah, you changed the question, but this does not fix the explanations. I am still confused about this one.
1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
Pour 3l water into the 5l jug, pour the 3l jug full, pour 2l from it into the 5l jug (till it is full), empty the 5l jug, pour the remaining 1l into it, pour another 3l into it.
2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
I don't think so. You need ceil(log2(10)) measurements even [...] Goddamn why did I read the spoilers.
2.b. Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Same as with 2., except you need a superincreasing sequence of number of coins from the bags, so you can't confuse weights with sums of weights. 1a+2b+4c+8d will do nicely.
3. (accessible) You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
Easy, start both hourglasses at the same time. Flip the 7 minute one at the 7th minute. Flip the 7 minute one again at the 11th minute. You have 4 minutes till it runs out.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
5. (my favorite ) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Not necessarily. Based on area alone, they could. But since 4 disk of radius 1/2 can't cover 1 disk of radius 1, they might miss some rectangles as well. Can't provide solid proof. Hehe nice solution by the others.
God I hate that problem. Took me a whole day to understand it, even though I knew the answer. In the queen + 100 wives + 100 husbands version, every husband dies at the 100th day.
7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
The trick here is we are comparing THREE groups of coins at once. Two on the scale, one in our hands. It is quite clear which group the counterfeit coin is in, be the scale balanced or unbalanced. So we take 3 coins in our hands and compare 3 coins against 3 coins. We know which group the counterfeit coin is in. We take 1 coin in our hands and compare 1 coin against 1 coin. We know which one is the counterfeit.
8. (Day9 wants you to do this one) On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes? Note: the string possibly burns unevenly.
Light one string on one end, the other string on both ends. Once the other strings burns out, light the other end of the first string as well.
9. (if you know what prime numbers are) When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
It sure as hell can't be divisible by 2, 3 or 5. That leaves 7, 11, 13, 17, 19, 23, 29 - all primes.
10. On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
All of them see the hats of those later in the line, the first one in line sees them all, except his. He communicates the parity of the sum of the binary representation of hats (0 or 1) by saying the respective color of the parity (it doesn't matter which color is 0 and which is 1). The guy next in the line can deduce his hat color based on this parity and what he sees. The others can deduce their hat color from the parity, and what others have said before him. Only the first in the line might die, the others surely live. Unless one of them fucks up
11. Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
No idea.
12. (Monty Hall?) On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes, this is a very famous problem. You should switch, your chances increase to 2/3 from 1/3 or so.
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
Yes, they can be. And it is unique. It's simply the binary representation of the numbers.
14. On April 10 2011 02:14 ghrur wrote: Question: What is the maximum number of times 10 lines can intersect in the same plane? paraphrased: What is the maximum number of intersection points between 10 distinct lines on a plane?
Looks like it is always possible to construct n lines so every line intersect the others. So it is 10 choose 2 = 45.
15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
In other words: Can you select 15 integers from 100 distinct integers such that all of them are in the same equivalence class mod 7? The answer is yes, if there are only 14 integers from each equivalence class, that is only 14 * 7 = 98, so it is guaranteed there are at least one equivalence class with 15 integers falling into it.
you are incorrect, but the puzzle is hard. Read my explanation if you are confused. Basically, there is a clever way to get around the "middle sister problem."
Yeah, the puzzle becomes obvious when you can ask the middle daughter something about the other daughters. Once you think of a question that never points to the middle daughter when asking the other daughters you win!
I am sorry, I still don't see, how there could be a question, that will evade the middle sister problem with a YES/NO question, since her answer will always be random, hence there can no be solution. I personally, can not think about
@cmpcmp Yeah, you changed the question, but this does not fix the explanations. I am still confused about this one.
I've heard this question before but I forgot the answer and I'm too lazy to think about it now but it has to do with asking one of the sisters if another will tell the truth or is more likely to tell the truth. I forgot.
Edit: my bad it's like the same thing so nevermind
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
"Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph."
Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks.
As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer
I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others.
Essentially, my solution picks a sister at random (which you must do) and asks her a question. The trick is that you NEVER pick the sister that you asked the question to. Thus, if you ask the middle sister the question it doesn't matter what she answers because you can't possibly pick her.
So: 1. Ask oldest sister: you pick the youngest sister (she told the truth) 2. Ask middle sister: she answers randomly and you pick the oldest sister 1/2 of the time and the youngest sister 1/2 of the time. 3. Ask the youngest sister: you pick the oldest sister (she lied about which one was younger)
10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
"Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph."
Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks.
As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer
I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others.
You are still completely wrong because brown eyed people would not have any indication that they can leave, ever.
Also, the whole premise of your solution being that we don't stick to the premise of the problem... That just sounds retarded. Hey, let's do 1+2 except let's assume that + isn't actually + because if we bend the rules a little it can be -. Oh wait, shall we enforce the RULE of + SOOOO STRICTLY? Jesus dude.
I (think) understand everybody's logic... and it makes sense if you follow those strict mathematical theorms.
What I don't understand is why they are concluding they have blue eyes when they can see other people with blue eyes. If everybody with blue eyes can see someone with blue eyes, nobody will ever leave the island.
I don't see how after 100 days that suddenly means they have blue eyes.
If the ferry takes somebody away every night... and on the first night after the guru speaks nobody leaves... does that not break the condition of somebody had to have left? Which means for 99 days you are counting 99 (or 100) blue eyes on the island. On the 100th day you realize, wow nobody is leaving.... I must have blue eyes? But wait, theres still 99 people with blue eyes here why aren't they leaving? What is so special about 100?
This is probably all jumbled up so don't be too hard on me... I can see why it works if you follow those definitions from the solution of the website.... But only if you assume those theorems are true. They have not given "proof" mind you. Something mathematical textbooks ALWAYS put in a coloured box after a theorem like that.
Case 1: You see 0 people with blue eyes. That's enough to determine your eye color, so you will leave on the first night(the words of the guru guarantee at least 1 person with blue eyes, so that would be you).
Case 2: You see 1 person with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 2.1. Your eyes aren't blue and the other person leaves on the first night(case 1 applies to him) Case 2.2. Your eyes are blue and the other person does not leave on the first night(case 2 applies to him). If your eye color wasn't blue he would've left, so you now know you have blue eyes(the same thing applies to him) and both of you leave on the second night.
Case 3: You see 2 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 3.1. Your eyes aren't blue and the 2 other people leave on the second night. Case 3.2. Your eyes are blue and the 2 other people don't leave on the second night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to them) and the three of you leave on the third night. ... Case 100: You see 99 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 100.1. Your eyes aren't blue and the 99 people with blue eyes will leave on the 99th night. Case 100.2. Your eyes are blue and the 99 other people don't leave on the 99th night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to all of them) and all 100 of you leave on the 100th night.
Essentially, my solution picks a sister at random (which you must do) and asks her a question. The trick is that you NEVER pick the sister that you asked the question to. Thus, if you ask the middle sister the question it doesn't matter what she answers because you can't possibly pick her.
So: 1. Ask oldest sister: you pick the youngest sister (she told the truth) 2. Ask middle sister: she answers randomly and you pick the oldest sister 1/2 of the time and the youngest sister 1/2 of the time. 3. Ask the youngest sister: you pick the oldest sister (she lied about which one was younger)
I finally see it. OMG!, I had to type it, to make sure it all works out. However you missed an important bit of explanation. + Show Spoiler +
You have to always skip the sister you ask the question to and choose the younger one from the remaining two, assuming the answer is correct.
So, all possible scenarios:
1. You asked the middle one. It doesn't matter. You win.
2. You asked the oldest, if THAT one is younger then the other and pointing at: (and that part is actually random) a. At the youngest. She will tell the true and tell you Yes. You pick her and win. b. At the middle. She will lie and say "No", you pick the youngest and win.
3. You asked the youngest, if THAT one is younger then the other and pointing at: a. At the oldest. She will lie and tell you Yes. You pick her and win. b. At the middle. She will lie and say "No", you pick the youngest and win.
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
"Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph."
Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks.
As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer
I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others.
You are still completely wrong because brown eyed people would not have any indication that they can leave, ever.
Also, the whole premise of your solution being that we don't stick to the premise of the problem... That just sounds retarded. Hey, let's do 1+2 except let's assume that + isn't actually + because if we bend the rules a little it can be -. Oh wait, shall we enforce the RULE of + SOOOO STRICTLY? Jesus dude.
You just being angry right now. Chill, dude. It's just riddles for which correct wording is quite important.
Also, regardless, how you solve, the problem, the moment all blue eyes leave, the guru will not confirm there are people with blue eyes anymore on the island, so everybody else will know they have brown eyes, hence they can leave.
However I would like to task you with my own version of this riddle. And this one is actually much harder:
There is an island where people are either blue-eyed or brown-eyed. There is no other person with any other color of his/her eyes. No exceptions!.
On this island there are no reflecting surface or anything that a person may learn his/hers eyes color, unless somebody else tells him/her. However telling/showing anybody the color of their eyes is strictly forbidden as a taboo and will not be violated under any circumstances.
Is it possible for all people on the island to figure out the color of their eyes? If so, provide an answer, please.
Just put 1 coin from each bag on the scale arranged such that you know which bag the coins came from, simply remove 1 coin at a time lol. I guess that wouldn't count as 1 weighing, though you only are putting something on the scale once so maybe..
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
Let's number the pirates A,B,C,D,E,F,G,H,I,J with A being the captain and so on.
It is ever gets to I to make the split, he will die automatically since J will disagree. So I will automatically agree with H, and so H can do whatever he wants, i.e. take all the money.
Thus, it is in I and J's advantage to agree with G as long as they get more than 0 (otherwise they might as well watch G get killed). Thus if it ever gets to G to make the split, he will be able to get away with this distribution (from G to J): 98-0-1-1. (H will get nothing because whatever offer G makes to him, he will disagree since it is in his advantage (H's) to disagree and become the one to make the split)
F can prevent this by offering this distribution (from F to J): 97-0-1-0-2 OR 97-0-1-2-0. From all steps here onwards, there is more than 1 choice, so the final distribution is not a unique solution. I will pick the former (97-0-1-0-2) in this step and pick an arbitrary choice for the following steps, which are as follows: A--B--C--D--E--F--G--H--I--J ----------------96--0--1--2--1--0 All the 0's disagree while all the nonzeros agree, since they will get more than they will if it goes to the next split.
------------96--0--1--2--0--0--1
--------95--0--1--2--0--1--1--0
----95--0--1--2--0--1--0--0--1
94--0--1--2--0--1--0--1--1--0.
So the captain can always get away with $94. The captain has many choices as to how to split the rest which I will not bother to list.
Edit: Ok, since apparently the split suggester's vote counts, my solution is invalid.
Edit2: Now under the assumption that the split suggester's vote counts, and using the same method, the distribution 96-0-1-0-1-0-1-0-1-0 is reached. This time, the solution is unique.
Convert any number into base 2, and you'll get the powers of 2 you need to summate. Since each base 2 number corresponds with a single base 10 number, each expression is unique
15 integers in any collection of 100 integers must have of the same equivalence class mod 7.
Assume there's only 14 of each class. 7 * 14 = 98 numbers total. Thus, we must add in two more numbers. Since each number must fall into one of the 7 equivalence classes, at least one class will have at least 15 numbers. Thus, we can pick those 15 numbers, and the terms of the problem will be satisfied
QED
Now for my own problem
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
English is not my mother tongue, so I apologize for some mathematical proofs that might be hard to follow.
1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol) + Show Spoiler +
Fill the 5 liter jug. Empty it into the 3 liter one. Empty the 3 liter jug in the pool. Empty the remainder of the 5 liter jug in the 3 liter jug. Fill the 5 liter jug. Fill the 3 liter jug with the 5 liter jug until it is full (that is, drop 1 liter from the 5 liter jug into the 3 liter jug). You have 4 liters in the 5 liter jug.
2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing? + Show Spoiler +
Place one bag on the scale, then another one. If the scales are even, you don't have the 9 grams bag. If they are not, you found it (ligher one). Rince and repeat, adding two additional bags each time'till you come to the point where you have the same number of bags on each side but the scales are uneven. The moment you add a pair of bags and the scales are uneven, the last one you added on the ligther side is the 9 grams bag.
3. (accessible) You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass. + Show Spoiler +
Start both hourglasses at the same time. Start boiling the egg when the 7 minutes hourglass ends. When the 11 minutes hourglass ends, reset it and wait until the end. Take the egg off the fire when this hourglass ends.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide? + Show Spoiler +
The hour hand goes 12 times slower than the minute hand. Therefore the next time they cross wil be between 1 and 2. Let x be the minute differential to 1 when the hands cross. Let's consider the circle has a size of "1". 1/12 + x/(12*60) = x/60 <=> x=60/11=~5.45. The hands will cross at 5.45 past one.
7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale? + Show Spoiler +
Yes, Place six coins on the scales If it is even, then put two of the remaining coins on the scales. If it is even, the counterfeit one is the one that is left. If it is not, it's the coin on the lighter side. If is not, then take the three coins from the lighter side of the scale and do what we would have done before with the three remaining coins.
9. (if you know what prime numbers are) When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not? + Show Spoiler +
Yes, Let p be a prime number greater than 32. There exists a unique (q,r) couple of integers that are so that p =30q+r
Suppose that r is not a prime number. Then there exists (a,b) couple of integers that are so that r=ab and a>1, b>1. What's more, a and b are odd, because p is odd (because any even number greater than 2 is not prime)
So, granted that ab is stricly smaller than 30, here are the possible situations : a=3,b=w/e, then p=3(10q+b) and p is not prime. That is absurd. a=5, b=w/e, then p=5(6q+b) and p is not prime. That is absurd. a=7, then b is smaller than 5 (because 5*7=35), so b=3 because b is odd and b is not equal to 1 and we have the same result.
Therefore, it is absurd that r is not a prime number. Hence, r is a prime number.
11. Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.) + Show Spoiler +
I don't think so. A 13-gon has a total of 13 corners. If you use parallelograms to tile it, you will end up having an even number of corners.
12. (Monty Hall?)
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes. The probability that you picked the correct door is 1/3. The probability that the correct door is the other one knowing that the 3rd door contained a goat is 1/2
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)? + Show Spoiler +
Let's prove the following property by induction: P(n):"For any integer p between 2^n and 2^(n+1), there exists an unique sum of distinct powers of 2 so that p is equal to that sum"
For n=0 1=2^0 2=2^1 Hence, P(1) is true.
Let n be an integer. Let's suppose that P(0),...,P(n) are true and let's prove P(n+1). Let p be an integer between 2^n and 2^(n+1).
If p=2^n the problem is solved.
Otherwise, p-2^n is an integer between 1 and 2^(n)(2-1)=2^n. Hence, there exists an unique expression as the one we are looking for, by induction.
I (think) understand everybody's logic... and it makes sense if you follow those strict mathematical theorms.
What I don't understand is why they are concluding they have blue eyes when they can see other people with blue eyes. If everybody with blue eyes can see someone with blue eyes, nobody will ever leave the island.
I don't see how after 100 days that suddenly means they have blue eyes.
If the ferry takes somebody away every night... and on the first night after the guru speaks nobody leaves... does that not break the condition of somebody had to have left? Which means for 99 days you are counting 99 (or 100) blue eyes on the island. On the 100th day you realize, wow nobody is leaving.... I must have blue eyes? But wait, theres still 99 people with blue eyes here why aren't they leaving? What is so special about 100?
This is probably all jumbled up so don't be too hard on me... I can see why it works if you follow those definitions from the solution of the website.... But only if you assume those theorems are true. They have not given "proof" mind you. Something mathematical textbooks ALWAYS put in a coloured box after a theorem like that.
Case 1: You see 0 people with blue eyes. That's enough to determine your eye color, so you will leave on the first night(the words of the guru guarantee at least 1 person with blue eyes, so that would be you).
Case 2: You see 1 person with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 2.1. Your eyes aren't blue and the other person leaves on the first night(case 1 applies to him) Case 2.2. Your eyes are blue and the other person does not leave on the first night(case 2 applies to him). If your eye color wasn't blue he would've left, so you now know you have blue eyes(the same thing applies to him) and both of you leave on the second night.
Case 3: You see 2 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 3.1. Your eyes aren't blue and the 2 other people leave on the second night. Case 3.2. Your eyes are blue and the 2 other people don't leave on the second night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to them) and the three of you leave on the third night. ... Case 100: You see 99 people with blue eyes. That's not enough to determine your eye color, so you don't leave. Case 100.1. Your eyes aren't blue and the 99 people with blue eyes will leave on the 99th night. Case 100.2. Your eyes are blue and the 99 other people don't leave on the 99th night. If your eye color wasn't blue they would've left, so you now know you have blue eyes(the same thing applies to all of them) and all 100 of you leave on the 100th night.
I'm still very unclear about the cases >3 (does not really matter how many other colours are around.)
I'm going to use I in place of the observer. (blue eyes)
Case 1---> 1 blue eyes; nobody else has blue eyes. I must have blue eyes and leave. Case 2 --->2 blue eyes; I see the guy with blue eyes did not leave, I must have blue eyes as well. Case 3 --->3 blue eyes; I see two other people with blue eyes, they must also see at least 1 set of blue eyes. I can see two sets of blue eyes, eliminating the possibility that they must see MORE than 1 set of blue eyes.
Why do I assume that those two people can see more than 1 set?
Thats the part I don't understand really I guess...
Scrap it, hit me like those two transports on Mythbusters and I'm the smart car.
I like logic puzzles more than math puzzles so... These are my two favorites :D
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
-----
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer realized he would never be able to make a stronger poison - That the pharmacist's poison would always be stronger then his own. So he instead makesa solution that has no poison in it at all, then before the contest starts drinks a poison weaker then the pharmacist's would be. This way he will survive having his weaker poison neutralized by the pharmacist's stronger poison and the pharmacist would die from his own poison. The pharmacist realizes this and instead of making a really strong poison makes a solution of no poison at all as well. So that when the contest comes, they both drink no poison at all, and since the treasurer drank the weak poison before, he dies from it, and the pharmacist stays alive, and the king gets no poison at all.
On April 10 2011 05:44 Joe12 wrote: A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Its a while since i heard this one, but im pretty sure the wording is correct..
HAHA by trying to prove, that this riddle isnt answerable in just 1 way, it hit me. 100 people drown. for every additional boat, there are 10 more people sinking with the original 10 boats.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
We can do this problem backwards. I will name it so the last person to go will be called the 1st and last is the first person to offer for simplicity. When it's one person, he gets it all. If there are two people, the 2nd gets it all If there are 3 people, then the 2nd person will always disagree the split, but the 1st will take anything he can get, so the 3rd gets 99, and the first gets 1. If there are 4 people, the 3rd will always disagree. The 4th must either offer the 2nd or the 1st more than the would have gotten in the above case, so 4th gets 99, 2nd gets 1. From here on will can assume the person that goes after the current will automatically disagree. If there are 5 people, he must offer the 2/3 of the first 3 more than they would have gotten. So, 5th gets 98, the 3rd gets 1, the 1st gets 1. We can continue this trend to get the answer.
EDIT:
On April 10 2011 05:44 Joe12 wrote: A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Its a while since i heard this one, but im pretty sure the wording is correct..
Let X be the number of boats and the number of men in each boat. Then we have x^2 people to begin with. We lost 10 lifeboats, each of which has x men, so we have x^2-10x. However, we then add 10 men in each of the remaining boats, of which there are (x-10), so we have x^2 -10x +10x-100, or 100 people drown.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
sry for really getting on your nerves today, but this is also wrong. just to give you 1 example of what you havent thought of: when 3 pirates are left, the 1. pirate will be able to get 99gold and give the last pirate 1 gold. the last pirate obv. knows that he cant get any gold from declining this offer (as you stated) so he will take the 1 gold.
ofc. this isnt the finals answer cause this way of thinking will continue to the higher first levels also.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
Let's number the pirates A,B,C,D,E,F,G,H,I,J with A being the captain and so on.
It is ever gets to I to make the split, he will die automatically since J will disagree. So I will automatically agree with H, and so H can do whatever he wants, i.e. take all the money.
Thus, it is in I and J's advantage to agree with G as long as they get more than 0 (otherwise they might as well watch G get killed). Thus if it ever gets to G to make the split, he will be able to get away with this distribution (from G to J): 98-0-1-1. (H will get nothing because whatever offer G makes to him, he will disagree since it is in his advantage (H's) to disagree and become the one to make the split)
F can prevent this by offering this distribution (from F to J): 97-0-1-0-2 OR 97-0-1-2-0. From all steps here onwards, there is more than 1 choice, so the final distribution is not a unique solution. I will pick the former (97-0-1-0-2) in this step and pick an arbitrary choice for the following steps, which are as follows: A--B--C--D--E--F--G--H--I--J ----------------96--0--1--2--1--0 All the 0's disagree while all the nonzeros agree, since they will get more than they will if it goes to the next split.
------------96--0--1--2--0--0--1
--------95--0--1--2--0--1--1--0
----95--0--1--2--0--1--0--0--1
94--0--1--2--0--1--0--1--1--0.
So the captain can always get away with $94. The captain has many choices as to how to split the rest which I will not bother to list.
Edit: Ok, since apparently the split suggester's vote counts, my solution is invalid.
Edit2: Now under the assumption that the split suggester's vote counts, and using the same method, the distribution 96-0-1-0-1-0-1-0-1-0 is reached. This time, the solution is unique.
Just read my solution for when split suggester's vote does not count, and understand how the logic works. Then apply the same method to get the answer in my "Edit2"
On April 10 2011 01:08 Slunk wrote: Question to #6 Is it correct to assume that the guru is completely useless and only built in as a distraction?
Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people
]But the guru only speaks once and tells something that everybody knows allready.
Two random people will stand next to each other. All other guys on the islands (save the guru - he already know his color (he DOES know he is the guru, since he executes the role)) will stand in a column in front of them.
1. First guy in the column will go to the first guys. 1a. If they have the the same eye color, he will go to one of their sides. Doesn't matter left or right. 1b. If they have different eyes colors he will go to the middle between them, hence there will be clear "middle" point between the half-row with blue eyes and half-row with brown eyes.
2. Second guy and so forth, save the last one will repeat step 1.
This way the one that goes in the "middle" point always sees where people in front of him divide by eye color and move between them, creating new such point.
3. The last guy will have to count the number of people with blue eyes and people with brown eyes in front of him (We shall assume he has not done so yet). The statement of the tasks clearly says:
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru.
By counting which group in front of him has 99 people, he will know to which group he belong and move to the side of corresponding half-row.
By doing so, all the rest 99 people will know the color of their eyes, as they will see the color of last person eyes when he moves. The other half-row will also know, they have the opposite color of the last man.
Well, now everybody knows the color of his (hers with the guru) and all 201 people will leave the island on the first night.
You might need to visualize this to understand how it works, but it's pretty simple and straightforward. It's actually harder to explain all the steps.
Do i get a virtual present now?
No, you are incredibly wrong.
Please, tell me where is the flaw in logic. I am not trying to troll, but rather asking politely.
If you disprove basing on the "no communication" rule, then in this case I will agree, that my solution it's not possible. However, I will stand by, that the rules does not forbid such interaction and my solution it's achievable - please, note that it does not requite and single person to say a word.
"Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph."
Communication is not limited to verbal communication. It means any sort of communication whatsoever. They include this so that people don't try to look for loopholes and avoid actually trying to think about legitimate solutions. Do they have to include all forms of communication in the conditions for you to understand this? No hand signaling, winking, systems of sorting people by eye color, pointing, grunting... That is what no communication means. Thanks.
As I said already, shall we enforce the "no communication" rule strictly, my solution it's not possible - in this case I have already provided the answer
I was asking if there is otherwise is any flaw in my logic, which allows everybody on the island to figure out the color of their eyes by simple grouping/ordering, without anybody actually communicating, but only taking action based on information from visually inspection others.
You are still completely wrong because brown eyed people would not have any indication that they can leave, ever.
Also, the whole premise of your solution being that we don't stick to the premise of the problem... That just sounds retarded. Hey, let's do 1+2 except let's assume that + isn't actually + because if we bend the rules a little it can be -. Oh wait, shall we enforce the RULE of + SOOOO STRICTLY? Jesus dude.
You just being angry right now. Chill, dude. It's just riddles for which correct wording is quite important.
Also, regardless, how you solve, the problem, the moment all blue eyes leave, the guru will not confirm there are people with blue eyes anymore on the island, so everybody else will know they have brown eyes, hence they can leave.
However I would like to task you with my own version of this riddle. And this one is actually much harder:
No, that's not correct. The Guru only speaks once, so he will never confirm that there are no more people with blue eyes on the island. Also there's the possibility that your eyes are not brown or blue. F.ex. your eyes could be red, you would never know since the Guru never speaks of people with red eye color.
The answer obviously lies with the Guru's statement. Since they all are perfect logicians and noone has left the island before the guru made his statement it changes everything. I don't know why though
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
Well, one fairly inefficient way to solve this problem is to designate counters, and use one of the switches to be a dummy switch that you switch to give no information, so the choices are basically switch or no switch on a single lightblub. The counter will just be the only person that flips the switch off, and other people would switch it on if and only if it is off and they have not been there before. It might also be possible to have multiple counters and a dynamic way of choosing counters. I don't think this is very efficient since it actually uses less possible signals then we are given.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
There is one problem I can see with your answer - Won't the pirate makign the spilt jsut split it up evenly with the number of other pirates he needs to get the vote? So first step instead of 11 to all he offers 25 to four, in order to try and get the 5 votes he needs. This would mean that the 7th pirate would offer all 100 gold to the eighth pirate, and both of them would vote yes the 7th to live the 8th because he would get all the money. If i read it wrong and the person who suggests the split doesn't get to vote then the same thing would happen with the 8th offering 100g to the ninth.
On April 10 2011 07:14 MusicalPulse wrote: In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer figured he had no chance to make the stronger poison, so he simply turned in something non-poisonous liquid (e.g. water), hoping the pharmacist would die from his own poison. The pharmacist realized this, so he made two poisons, and he either swallowed the weaker before the meeting, or the stronger after it. Nonetheless the king didn't necessarily get the strongest poison possible in either case.
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
We can do this problem backwards. I will name it so the last person to go will be called the 1st and last is the first person to offer for simplicity. When it's one person, he gets it all. If there are two people, the 2nd gets it all If there are 3 people, then the 2nd person will always disagree the split, but the 1st will take anything he can get, so the 3rd gets 99, and the first gets 1. If there are 4 people, the 3rd will always disagree. The 4th must either offer the 2nd or the 1st more than the would have gotten in the above case, so 4th gets 99, 2nd gets 1. From here on will can assume the person that goes after the current will automatically disagree. If there are 5 people, he must offer the 2/3 of the first 3 more than they would have gotten. So, 5th gets 98, the 3rd gets 1, the 1st gets 1. We can continue this trend to get the answer.
I will disagree and will change my answer, but for different reasons and more complicated logic:
Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
-------
When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
-------
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails.
--------
When there are 4 people: 4th will propose: 1 - 0, 2-0, 3 - 0, 4-100
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will stay alive and will try to make the most of the money. He will vote yes 2:2 split accepted.
---------
When there are 5 people: 5th will propose - irrelevant.
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will try to kill more and he know he can stay alive. He will vote No. 5th will vote yes, to stay alive. 2:3 - split denied.
----------
When there are 6 people: 6th will propose
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will can safely vote No, because he can safely kill more pirates. 4th will try to kill more - what he gets doesn't matter. He will vote No. 5th will vote yes, in oder to stay alive. 6th will try yo stay alive. He will vote Yes" 2:4- split denied.
At this point, there is no point to keep going onward. The final answer is there will be 4 pirates left, and 4th one (counted backwards) will get all the money.
On April 10 2011 07:59 LastPrime wrote: A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
Does the vote include the person making the split suggestion?
To satisfy condition #1, the captain will offer whatever he can to say alive (At the maximum he can offer its 11 pieces to everybody. Anything else is irrelevant). The rest of the crew minus the next in command can safely vote to fail, since condition #1 is true and they try to achieve condition #2 - get more money and condition #3 - kill more pirates.
The cycle repeats itself until there are only 2 people left. At this points all conditions are true.
Half of the crew will agree this is the split: 9th pirate will propose that 9th gets 100, 10th get 0 and the split is approved. The last person can no break this as the split simple gets approved with 50 per cent majority.
sry for really getting on your nerves today, but this is also wrong. just to give you 1 example of what you havent thought of: when 3 pirates are left, the 1. pirate will be able to get 99gold and give the last pirate 1 gold. the last pirate obv. knows that he cant get any gold from declining this offer (as you stated) so he will take the 1 gold.
ofc. this isnt the finals answer cause this way of thinking will continue to the higher first levels also.
ill will try an figure it out now
You are correct. Don't worry about nerves - as I said earlier today it's just riddles. However before I read your post, I already saw a flaw in my logic based on terr13 posts.
Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
-------
When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
-------
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails.
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When there are 4 people: 4th will propose: 1 - 0, 2-0, 3 - 0, 4-100
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will stay alive and will try to make the most of the money. He will vote yes 2:2 split accepted.
---------
When there are 5 people: 5th will propose - irrelevant.
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will try to stay alive, because if the vote fails, he will be killed. He will vote yes. 4th will try to kill more and he know he can stay alive. He will vote No. 5th will vote yes, to stay alive. 2:3 - split denied.
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When there are 6 people: 6th will propose
1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd will can safely vote No, because he can safely kill more pirates. 4th will try to kill more - what he gets doesn't matter. He will vote No. 5th will vote yes, in oder to stay alive. 6th will try yo stay alive. He will vote Yes" 2:4- split denied.
At this point, there is no point to keep going onward. The final answer is there will be 4 pirates left, and 4th one (counted backwards) will get all the money.
Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
-------
When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
-------
Agreed.
[quote]
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails. [\quote]
Here is where your logic fails to hold. 3rd person's offer is not irrelevant. If he offers 1 to 1st person, the first person will accept, because it is better than getting 0 from 2nd person's offer.
[QUOTE]On April 10 2011 08:10 LastPrime wrote: [quote]Lets work it backwards, but to make sure that all conditions will be properly executed. When there is 1 person - he gets all.
-------
When there are 2 people. 1st get 0, 2nd gets 100. 1st one votes - No, 2nd votes Yes. 1:1 votes - split accepted!
------- [/quote]
Agreed.
[quote]
When there are 3 people, 3rd person will offer - irrelevant. 1st knows he can't get anything, so he will for the vote to fail, in order to kill more pirates. 2nd knows he can get 100 and stay alive, so he will vote to fail, in order to satisfy last condition - kill more. 3rd person - what ever he offers doesn't matter. 1:2, split fails. [\quote]
Here is where your logic fails to hold. 3rd person's offer is not irrelevant. If he offers 1 to 1st person, the first person will accept, because it is better than getting 0 from 2nd person's offer. [/QUOTE]
You are absolutely correct. Let me work this again and see where I arrive. Will post in edit.
On April 10 2011 07:59 LastPrime wrote: A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
1. (accessible to everyone) you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
Fill in the 5 liter, then spill it into the 3 liter. When the 3 liter is full, you have 2 liters left in the big one. Do it twice, and you got 4 liters.
2. Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Take 1 from the first bag, 2 from the second bag, 3 from the third, etc... Multiply the number of grams missing (the 55 (1+2+3+...+10=55) coins should weigh 55 grams if they were all real) by 10 and you have the number of the bag with fake coins.
2.b. Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Do the same thing with 1/2/4/8 coins from bags #1/2/3/4.
3. (accessible) You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
Flip them both. When 7 is empty, flip the 7. 7 minutes have gone by. When 11 is empty, 11 minutes have gone by, and 4 minutes since you flipped the 7. Flip the 7 again, and wait for it to be empty. 15 minutes have gone by.
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
It's easy to find that it's around 1h05. To find the exact time, lets call x the angle that the hands do with the 12. Hour and Minutes are on the x. "60+x minutes" has to be equal to "x hours". "x hours" is x*12/360 hours, or x*12/6 minutes, or 2x minutes (for example, 90° = 180 minutes = 3 hours). "x minutes" is x*60/360 minutes or x/6 minutes (for example 90° = 15 minutes). So we need to have 60+x/6 = 2x, which gives 11x/6 = 60 or x = 360/11 degrees. Both hands are on 360/11 degrees, which means it's 1:05:05
5. (my favorite :p) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
All 100 guys with blue eyes will leave on the 100th day. Let's assume there's only one guy with blue eyes. He'll look around him and see no one has blue eyes, but he knows someone does, so he knows it has to be him, so he leaves on the first day. Now, if there was 2 guys with blue eyes, each of them would think "This dude is the only one I can see who has blue eyes. If I don't have blue eyes, he doesn't see anyone with blue eyes, so he'll leave. Let's see if he's still here tomorrow..." The next day, the other dude's still here, so it means I too have blue eyes. The other guy thinks the exact same thing, and we both leave on the 2nd day. [...] Now, with 100 guys with blue eyes, each one of them will think "I see 99 guys with blue eyes. If I don't have blue eyes, they'll all leave on the 99th day. Let's see if that happens. On the 100th day, the other 99 guys are still here, which means I too have blue eyes. The other 99 guys think the same thing, we all figure out we have blue eyes, and all leave on the 100th day.
7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
Split them in 3 piles of 3. Weigh one of the piles against the other. 2 possibilities : - it's balanced : the lighter coin is in the 3rd pile - it's not balanced : the lighter coin is on the lighter side Either way, we now know in which pile of 3 the lighter coin is. Weigh one coin of this pile against another of this pile. - it's balanced : the 3rd coin is lighter. - it's not balanced : we found the lighter coin.
8. (Day9 wants you to do this one)
On April 10 2011 00:29 Munk-E wrote: Of course I must add, If you have 2 pieces of string that when you light in fire take an hour to burn how do you measure 45 minutes?
Light one on one end, and the other on both hands. When the one you lighted on both end is done burning, 30 minutes have gone by. Light the other hand of the first one, when it's done burning, 15 more minutes have gone by, total 45 minutes.
9. (if you know what prime numbers are) When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
Lets call A our prime number greater than 32. R < 30. Lets suppose R is not a prime number.
10.
On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
The first vizir makes 2 steps forward. The second vizir comes close to the other. Now the third vizir has 2 options : - if the first 2 have different colors, he goes between them - if the first 2 have the same color hats, he goes to either end of this new line. The 4th vizir does the same : - if all 3 hats he sees in the new line are the same color, he goes to one end of the line - if there are hats of both colors, he goes between the red and blue that are next to one another. In the end, if they all do that, all reds will be on one side, all blues on the other. They will all announce their color starting from the most left, then the most right, then the second-most left, then the second most-right. Only one will be wrong (or maybe they'll all be right, if they're lucky).
11. Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
Too complicated for 1:10 am, can't be bothered :D
12. (Monty Hall?)
On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Yes you should switch. If you picked the door with the car first, switching means losing and not switching means winning. If you picked the door with the goat first, switching means winning and not switching means losing. But the second situation happens twice more often, which means you double your chance of winning by switching (2/3 vs 1/3).
13. Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
45? Each line can cross every other once, and I don't think there's a reason why some intersection points should be common, so for each line there is 9 intersection points. But you must not count twice the point of intersection of line A and B and line B and A, so half of 10*9. It seems too easy, maybe there's a catch.
15. Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
I'll think about it tomorrow :D Too late and too hard
1 person /w blue --> understandable 2 people /w blue---> understandable >3 people /w blue ---> ????
The 1 person example; no other blue eyes, obviously. The 2 people example; the 2nd person knows the first person can see a 2nd set of blue eyes. easy peazy. The 3 people example; the 3rd person knows the other two can see each others eyes. Why would anybody leave?
If there's you, B(lue)1, B(lue)2, B(lue)3. Now there is 2 cases, either you have blue eyes or you don't. If you don't have blue eyes the only people with blue eyes that B2 will see is B1 and B3. B2 knows that if he didn't have blue eyes B1 and B3 would have left on the second night. If B1 and B3 doesn't leave on the second night B2 knows that he must have blue eyes aswell. And if B1, B2 and B3 doesn't leave on the third night then you know that you must be blue eyed. Using this logic you can figure it out.
The key is not thinking that they see eachother, instead take it in steps thinking, "If I don't have blue eyes, B1 only sees 2 people with blue eyes" and then you put yourself in the perspective of B1 with the assumption that you do not have blue eyes which would lead to him thinking "If I don't have blue eyes, B2 only sees 1 person with blue eyes which would make him leave on the second night"
10. On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
All of them see the hats of those later in the line, the first one in line sees them all, except his. He communicates the parity of the sum of the binary representation of hats (0 or 1) by saying the respective color of the parity (it doesn't matter which color is 0 and which is 1). The guy next in the line can deduce his hat color based on this parity and what he sees. The others can deduce their hat color from the parity, and what others have said before him. Only the first in the line might die, the others surely live. Unless one of them fucks up
On April 10 2011 00:05 stepover12 wrote: 7. (accessible to everyone) Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
I'll add more when I can.
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
On April 10 2011 08:43 Darkren wrote: 4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
10.) Easier than you think I had this in a leadership class. + Show Spoiler +
You will not be able to see the hat on your head. Is the statement you should focus on, if you are smart enough you'd just take the hat off your head so you can see it. These riddles usually go with something so obvious that you feel a little wetarded when you have to solve it.
Reminds me on "How fast can everyone touch the ball trick?" Our class of about 30 students all touched it in about 1 seconds flat after we figured it out, first we were passing the ball to each other got that down to about 20 something seconds, then we tried getting closer, got it down more, then we tried just handing it off in a circle got it down faster. Then we all put a finger on the ball and let go at the same time.
It just shows there is more than one solution, but which is best? In other words a lot of the solutions weren't wrong, and a lot weren't right, the ones where you have to sacrifice someone is just silly =P Nobody needed to die! just take the hat off and put it on your hand and say the color
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
I always forget the correct solution to this particular puzzle and instead I try to escort the 10-minute one with the 1-minute one, which leads to an incorrect solution -_-
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
The 2$ in the boy's pocket was paid by the men, it is already in the 3*9$ they paid. The missing 3$ was given back to the men, 3*9$+3$=30$. Deceptive puzzle, but can't be misunderstood when read online.
After a grueling process of keeping track of every possible offer each pirate member can have for every split, I've finally come to a conclusion. So working backwards, we start with the case with 2 pirates left. The one doing the splitting will be labeled as pirate#2, with the one lower in rank being pirate#1 (so the rest will be labeled in accordance to their ranks with the captain being pirate#10. As it has been discussed before, #2 has the power to split everything to himself, since his vote is enough as the majority. so:
Split1: #2-100, #1-0
From here a lot of reasoning will be cumulative and dependent upon the possible offers of previous splits. So now when #3 is doing the splitting, he can easily persuade #1 to agree with him with one coin, since it's better than getting 0 with #2 splitting. #2 will be inclined to always disagree until it is his turn to split, but his vote does not matter. #3 and #1 are enough as the majority. So:
Split2: #3-99, #2-0, #1-1
So #4 has to try and persuade at least 1 person to his side yet again like with #3. #2 and #3 are going to disagree no matter what since they stand to benefit much greater when it is their turn. So similarly with #3, he can just offer #1 a bit more than what #3 will have to offer.
Split3: #4-98, #3-0, #2-0, #1-2
Now things change up. #5 is in a bad position as #1 isn't enough for majority. In fact, #4, #3, and #2 will benefit tremendously from their splits only, so they'll never agree. so #5 is destined to die.
Split4: #5 will die
Things change greatly again. #6 is in a much better position than #5, since he can offer #5 0 as #5 will still have his life, and #1 a bit more than what #3 would have offered for majority!
Split5: #6-97, #5-0, #4-0, #3-0, #2-0, #1-3
Now like #5, #7 is also screwed since he can not ever get #6, #4, #3, or #2 to agree. He'll never reach majority and die.
Split6: #7-dies.
Do you see the pattern now? The odd ranks greater than 3 can not pull majority no matter what. They will die when they try to split. The even ranks can abuse this, offer minimum to those destined to die and #1 a bit more than what they would get so far. So moving along, #5 accepts 1 as it's now the best offer he can get after just his life in the previous offer. #7's new best offer is simply his life. This pattern will repeat itself until it's the captain doing the splitting.
The end result is one cunning captain with most of the treasure for himself, and the majority formed by #9, #7, #5, and #1 sucking it up and agreeing since there's absolutely no better possible outcome for them unless they agree.
One important possible flaw: I'm assuming the pirates will only accept the offer if it's GREATER than any possible future offer, not equal to. If the pirates are willing to agree even if future offers are the exact same, then the captain would just end up with all 100 and still get his majority to agree.
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
Wow, beautiful!
That logic doesnt work because the younger sister could lie and tell u the middle one is the youngest. U could also ask the older sister if the middle one is younger and than her and she will answer u yes.
U could ask do u think lying to someone is ok? If u pick the younger one she will tell u no(lie) If u pick the older one she will tell u no If u pick the middle one she will answer yes because she lies sometimes and tell the truth other times
U then know if she answers yes that she is the middle sister and if she answers no u marrie her
hmm in the pirate puzzle do we assume that the governing 2 rules are money and life? or money and kills? because for example: if there are 3 pirates left let's call them A B and C with A making the call the split like this should work A9 B:0 C:1. my reason for this is that A wants to keep as much of it as possible to himself and since he needs another vote to survive he gives the pirate with no chances of gaining money if he dies a piece of the loot in order to get the needed votes, i dont think it will ever come to only 2 remaining pirates.
[QUOTE]On April 10 2011 09:41 Frigo wrote: [QUOTE]On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint? [/quote] You sure this is possible?
U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
[QUOTE]On April 10 2011 10:39 Darkren wrote: [QUOTE]On April 10 2011 09:41 Frigo wrote: [QUOTE]On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint? [/quote] You sure this is possible?
U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it[/QUOTE]
If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs...
Pour 5 liters into 5 liter jug. Fill up the 3 liter jug using water from the 5 liter jug. You have 2 liters of water in the 5 liter jug now. Pour out the 3 liter jug. Pour the 2 liters of water in the 5 liter jug into the 3 liter jug. Fill the 5 liter jug again. Fill the 3 liter jug with water from the 5 liter jug. The 5 liter jug has 4 liters in it.
On April 10 2011 01:54 Kazius wrote: Ohh! I love these. 1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this):
You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door
[QUOTE]On April 10 2011 10:44 ixi.genocide wrote: [QUOTE]On April 10 2011 10:39 Darkren wrote: [QUOTE]On April 10 2011 09:41 Frigo wrote: [QUOTE]On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint? [/quote] You sure this is possible?
U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it[/QUOTE]
If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs... [/QUOTE]
No it's possible, i gave u the first part of the solution, this isn't a 2 min riddle u have to think on it.
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
U start by balancing 4 to 4 if it is equal u have 4 left and in those 4 is the coin u are looking for. U then balance 2 of those coins with 2 good ones, if it is not balanced u know that in those 2 is the bad one, then u take one of them and balanced with a good one. if it is equal u know the last one is the bad coin and if it is unbalanced u know that one is the bad coin. If on the second balance the 2 pair of coins are balanced u know the last 2 are the bad ones and u repeat a few steps up. Now comes the hard part if u have it unbalanced at the start 4 to 4
If u read the spoiler i just give a small part of the solution and u see how long it is. It is not an easy problem think about it
If the coin is in 1 of the 4 that you are weighting then you can only get down to 50% in 3 weighs...
No it's possible, i gave u the first part of the solution, this isn't a 2 min riddle u have to think on it.[/QUOTE]
Ur getting caught in the problem, multiplication and division alwais take the priority, the equation can be rewroten (48/2)(9+3) which equals 24 X 12 = 288, ur calculator can't see these errors
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B from the opposite side is heavy. If they do not balance, the lighter one is the incorrect weight.
Ur getting caught in the problem, multiplication and division alwais take the priority, the equation can be rewroten (48/2)(9+3) which equals 24 X 12 = 288, ur calculator can't see these errors
12 coins, 1 of which is either heavy or light EEEEEEEEEEEE (24 possibilities), measure 4v4 they balance, the 4 others are either heavy or light ___EEEE (8 possibilities), measure 3 versus 3 legit ___ they balance, the one you didn't measure is either heavy or light ______E (2 possibilities), measure it versus 1 legit ___ they don't, you have 3 heavy or 3 light ______hhh or lll (3 possibilities), compare 2 ______ they balance, the third is the fake ______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light ___hhhhllll (8 possibilities), measure hhl versus hhl ___ they balance, one of the 2 lights you didn't measure is fake ______ll (2 possibilities), measure versus each other ___ they don't, one of the 2 heavies was heavy or the opposite light was light ______hhl (3 possibilities), measure the heavies versus each other ______ they balance, the light was light ______ they don't, the heavy one is heavy
With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above
12 coins, 1 of which is either heavy or light EEEEEEEEEEEE (24 possibilities), measure 4v4 they balance, the 4 others are either heavy or light ___EEEE (8 possibilities), measure 3 versus 3 legit ___ they balance, the one you didn't measure is either heavy or light ______E (2 possibilities), measure it versus 1 legit ___ they don't, you have 3 heavy or 3 light ______hhh or lll (3 possibilities), compare 2 ______ they balance, the third is the fake ______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light ___hhhhllll (8 possibilities), measure hhl versus hhl ___ they balance, one of the 2 lights you didn't measure is fake ______ll (2 possibilities), measure versus each other ___ they don't, one of the 2 heavies was heavy or the opposite light was light ______hhl (3 possibilities), measure the heavies versus each other ______ they balance, the light was light ______ they don't, the heavy one is heavy
With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
12 coins, 1 of which is either heavy or light EEEEEEEEEEEE (24 possibilities), measure 4v4 they balance, the 4 others are either heavy or light ___EEEE (8 possibilities), measure 3 versus 3 legit ___ they balance, the one you didn't measure is either heavy or light ______E (2 possibilities), measure it versus 1 legit ___ they don't, you have 3 heavy or 3 light ______hhh or lll (3 possibilities), compare 2 ______ they balance, the third is the fake ______ they don't, the one that matches heavy or light is the fake
they don't, one of the 4 is heavy or one of the other four is light ___hhhhllll (8 possibilities), measure hhl versus hhl ___ they balance, one of the 2 lights you didn't measure is fake ______ll (2 possibilities), measure versus each other ___ they don't, one of the 2 heavies was heavy or the opposite light was light ______hhl (3 possibilities), measure the heavies versus each other ______ they balance, the light was light ______ they don't, the heavy one is heavy
With the 12 coin problem you can do better than just finding which is counterfeit, you should also be able to tell whether it is heavy or light.
I don't think it is even possible to find a solution WITHOUT telling if the counterfeit is heavy or light.
On April 10 2011 00:05 stepover12 wrote: 19.(pirates, arrrr)+ Show Spoiler +
On April 10 2011 06:32 tomnov wrote: 10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order + Show Spoiler +
* a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer
how do they split the money and how many pirates die?
I'll add more when I can.
Ive had my math teacher ask me this riddle but with the twist how much has the first pirate offer to not be killed and make the most money. Apparently he can make in the range of 30$-40$.
Ive not yet solved it any imput would be appreciated
Ur getting caught in the problem, multiplication and division alwais take the priority, the equation can be rewroten (48/2)(9+3) which equals 24 X 12 = 288, ur calculator can't see these errors
I can't see this one working. The problem is reduced to 8 coins of which 1 is counterfeit, with 4 real coins to help us and 2 weighings remaining. Unless I am missing something, here is how I would proceed to find the counterfeit: Weigh 4 of the suspected coins against the 4 real coins (or two suspected coins on each side). Regardless of the result we're now down to 4 possible counterfeits and 1 weighing. Like the "hint" that was posted earlier says, this requires 2 weighings to solve from here. I'd really like an explanation for this one...
With the way I was taught math: 2. This is more of an argument starter than an actual math problem since a mathematician would have the good sense to include another parenthesis.
I guess I'll pick this one you label as your favorite.
5.) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course. + Show Spoiler +
I don't see the point in them being disks, but anyways lets convert them to squares since circles are dumb and give the same results. Radius of 1 so they are 2x2 squares (4 units squared is their area). It's completely covered, so it must have an area of 4*25 = 100 units squared.
Now we got new smaller disks with a radius of 1/2 meaning they are equivalently 1x1 squares. So the area of 100 1x1 squares is 100 units squared. Same as last time.
So the answer is an obvious yes, 100*1 is indeed the same as 25*4.
My problem with these kind of math/brain teaser questions is they normally rely on you having to assume things because if they actually give you all the relevant information like in #5, they are too incredibly easy.
26. suppose you have a chess board with 2 opposite corners cut out as in picture there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
On April 10 2011 13:01 Befree wrote: I guess I'll pick this one you label as your favorite.
5.) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course. + Show Spoiler +
I don't see the point in them being disks, but anyways lets convert them to squares since circles are dumb and give the same results. Radius of 1 so they are 2x2 squares (4 units squared is their area). It's completely covered, so it must have an area of 4*25 = 100 units squared.
Now we got new smaller disks with a radius of 1/2 meaning they are equivalently 1x1 squares. So the area of 100 1x1 squares is 100 units squared. Same as last time.
So the answer is an obvious yes, 100*1 is indeed the same as 25*4.
My problem with these kind of math/brain teaser questions is they normally rely on you having to assume things because if they actually give you all the relevant information like in #5, they are too incredibly easy.
As I said:
On April 10 2011 00:46 stepover12 wrote: Ahh, for number 5, showing that the area is enough to cover may suggest the answer is yes. But it is not a definite proof that there is a way to cover the rectangle with those smaller disks.
There is an elegant solution without having to make any other assumption. Hint: + Show Spoiler +
divide and conquer.
solution for 5 (challenge yourself, think about the hint before looking down below): + Show Spoiler +
divide the rectangle into 4 smaller rectangles similar to the original. Each of these small rectangle can be covered by 25 small disks (of radius 1/2) the same way that 25 disks of radius 1 can cover the big rectangle. Putting them together, we get 100 small disks can cover the whole big rectangle. The answer is proven to be yes.
I can't see this one working. The problem is reduced to 8 coins of which 1 is counterfeit, with 4 real coins to help us and 2 weighings remaining. Unless I am missing something, here is how I would proceed to find the counterfeit: Weigh 4 of the suspected coins against the 4 real coins (or two suspected coins on each side). Regardless of the result we're now down to 4 possible counterfeits and 1 weighing. Like the "hint" that was posted earlier says, this requires 2 weighings to solve from here. I'd really like an explanation for this one...
With the way I was taught math: 2. This is more of an argument starter than an actual math problem since a mathematician would have the good sense to include another parenthesis.
You don't actually have 8 coins that contain a counterfeit with 2 weighings left. You have 12 coins left, and you know which 4 (minimum, assuming the first weighing is unbalanced) or 8 (maximum, assuming the first weighing is balanced) are normal. You can use this to force certain balance.
I'm pretty sure it is impossible. You must use shapes of 1x2 to cover a board that can be subdivided into 3 smaller grids of 1x7, 1x7, and 6x8. No matter how you place the tiles, you will eventually be in a situation where you must cover a grid of 1x7 squares with 1x2 tiles. This will prove to be impossible.
hint for people solving the 12 coin problem: drawing a table of probability will help a ton since the answer might require you to mix up the coin to predict the out come a little bit. Anyone who could solve this in their head must either hv heard of this b4, google it or is a freaking genius.
it took me a full night back in grade 9th just to solve that problem with a whole bunch of logic math books surrounding.
You don't actually have 8 coins that contain a counterfeit with 2 weighings left. You have 12 coins left, and you know which 4 (minimum, assuming the first weighing is unbalanced) or 8 (maximum, assuming the first weighing is balanced) are normal. You can use this to force certain balance.
Since a hint was posted earlier covering the easier scenario where the initial weighing was balanced, thus reducing the possible counterfeits to 4, I'm assuming the rest of the problem is, indeed, 8 possible counterfeits with 2 weighings remaining. Are you saying I'm already in the wrong at this point? For NB: The thought of bringing probability into a logic riddle is... worrying!
On April 10 2011 09:22 GertHeart wrote: 10.) Easier than you think I had this in a leadership class. + Show Spoiler +
You will not be able to see the hat on your head. Is the statement you should focus on, if you are smart enough you'd just take the hat off your head so you can see it. These riddles usually go with something so obvious that you feel a little wetarded when you have to solve it.
Reminds me on "How fast can everyone touch the ball trick?" Our class of about 30 students all touched it in about 1 seconds flat after we figured it out, first we were passing the ball to each other got that down to about 20 something seconds, then we tried getting closer, got it down more, then we tried just handing it off in a circle got it down faster. Then we all put a finger on the ball and let go at the same time.
It just shows there is more than one solution, but which is best? In other words a lot of the solutions weren't wrong, and a lot weren't right, the ones where you have to sacrifice someone is just silly =P Nobody needed to die! just take the hat off and put it on your hand and say the color
No, there is no trick. These are not stupid scenarios with an easy trick that makes you say "Oh, I should have thought of that." These are legit problems that actually require effort to solve. The answer to 10 is just modulo arithmetic.
The 12 coins problem solution is rather convoluted if I recall correctly, and it actually takes a while to sit down and write it out, or maybe my solution was not quite as elegant.
Separate the balls into 3 categories 4 A's, 4 B's, 4 C's. Weight AAAA vs BBBB. If they are equal, then the solution is simple. Say that AAAA> BBBB. Then we can weigh ABC vs BAB, and call this group 1 vs group 2. If ABC< BAB, then either B1 is lighter or A2 is heavier, and the answer follows. If they are equal, then either the two A's that weren't weighed are heavier, or the 1 B that wasn't weighted is lighter. If ABC>BAB, then either A1 is heavier, or either of the B2's are lighter, which is similar to the case above. In both cases we have 2 balls that are lighter or heavier, and one ball that is the opposite. Without loss of generality say AA are heavier, or B is lighter, and the rest of the balls are normal, denoted N. Then we can weigh AB vs NN. If AB is lighter, B is counterfeit. If AB is heavier, then that A is counterfeit. If AB is equal weight, then the one A that was not weighed is the counterfeit.
There is no way to cover the chess board with dominos. Each domino will cover one black and one white space, but the corners removed were both black. This has left 32 white spaces and 30 black spaces, so at best there will be two white spaces remaining after using 30 dominos.
On April 10 2011 13:10 NB wrote: hint for people solving the 12 coin problem: drawing a table of probability will help a ton since the answer might require you to mix up the coin to predict the out come a little bit. Anyone who could solve this in their head must either hv heard of this b4, google it or is a freaking genius.
it took me a full night back in grade 9th just to solve that problem with a whole bunch of logic math books surrounding.
I managed to solve the problem in my head but it took me a 2-3 hours doing it with a friend.
Ill give an additional hint u have three things that a balance tells u. Which sides weights more, how many coins are on each side and if u switch side u can see which coins have an effect and which have none.
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above
As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing)
Edit for clarification:
That is to say, if I arranged them using L for possible light, and H for possible heavy, the scale for the second weighing would look like LLH/LLH. If the left side were lighter, that would mean either one of the 2 Ls on the left was actually light, or the one H on the right was heavy, obviously the reverse is true if the right side registered lighter...bit difficult to express this in written form
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above
As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing)
Thats the whole problem u don't know that the B coins where heavier in the first weigting it could have been that or it could have been that the A coins were lighter
On April 10 2011 08:43 Darkren wrote: Ive got a couple love those things
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
You sure this is possible?
I believe this approach will work...unless someone can point out the flaw:
Place 3 coins on each side, set the rest aside -> Whether they balance or not, this leaves you with 6 coins with 2 weighs remaining, as well as 6 coins that you know are real, thus the proper weight. Of the 6 unknown, place 2 on each side of the scale -> Will leave you with 2 coins, with 1 remaining, and 10 that you know are the proper weight. Place one of the remaining coins on one side, with one of the known coins on the other. -> Either the unknown coin will be heavier/lighter, or will balance, meaning the last remaining coin is the counterfeit.
This does not work u are supposing that on the second balance u take 2 on 2 that u will take the bad coin. If it balances out u are screwed. Read one of my posts above and look at the start of the problem and continue from there
K, think I have it this time, heh, realized my earlier mistake and tried to ninja edit it right as you quoted :p
Divide 4/4/4 and place them onto the scale, if they balance it's quite easy as you posted earlier, so we'll address if it does not balance. I'll label the side that's lighter as A and the side that's heavier as B for ease, with those known as K.
Set up the scales for 2A 1B/2A 1B. If they balance, weigh the 2 Bs that are not on the scales against each other, the heavier is the incorrect weight. If they do not balance, weigh the 2 As from the lighter side against each other.
If these 2 balance, the B was heavy. If they do not balance, the lighter one is the incorrect weight.
On your first part if they balance and u weight the 2 B together how do u know the heavier one is the incorect one? U don't know if it is lighter or heavier so it could be any of those 2 U also do not know which As to weight eachother for the same reason as above
As the B coins were those that were heavier in the first weighing, that means that either one of the 4 A coins are light or one of the 4 B coins are heavy. By splitting them as I did, all of the possible 'light' coins were on the scale, while the 2 B that were considered if those balanced could only be unbalanced if one was heavier than it should be, else the initial weighing would have been different (heavy/light split based on observation from the first unbalanced weighing)
Thats the whole problem u don't know that the B coins where heavier in the first weigting it could have been that or it could have been that the A coins were lighter
But I do know it has to be one or the other, as otherwise that's the reason for the scales to have been unbalanced (if the left side rises up while the right sinks, for example, then either one coin from the left, or A set of coins, is light, or one coin from the right, or B set of coins is heavy, but not both.
The direction of the imbalance in the first weighing indicates which coins could possibly be light, or which ones could possibly be heavy, as you couldn't have a 'B' coin that's lighter than it should be, else the scale would have tipped in the other direction in the first weighing. This result of that weighing is how I label them as A and B, and supports the conclusions I'm making in the next weighing.
Possible I may not be explaining myself adequately, or may not be understanding what you're saying, heh =)
Edit: Just realized I mistyped my 3rd weighing conclusion...If you weigh 2 A's against each other, and they balance, the B you conclude is heavy was the B on the opposite side...managed to not realize i didnt type that correctly, gg me ><...updating
After having looked this riddle up, a few notes... + Show Spoiler +
The original question, as well as some of the hints given in this thread, were less than helpful. The real riddle demands that you determine if the fake coin is heavier or lighter, which actually makes the riddle easier. Nice try, Everhate. Your idea was creative, but it fails because you are only looking for the coin with the odd weight instead of... conducting a more thorough investigation!
On April 10 2011 15:52 eekon wrote: After having looked this riddle up, a few notes... + Show Spoiler +
The original question, as well as some of the hints given in this thread, were less than helpful. The real riddle demands that you determine if the fake coin is heavier or lighter, which actually makes the riddle easier. Nice try, Everhate. Your idea was creative, but it fails because you are only looking for the coin with the odd weight instead of... conducting a more thorough investigation!
Unless I managed to overlook something in the process, I believe, in each of the conclusions, which coin is indicated will also tell whether it is heavier or lighter than the rest, based on which stack it came from. That was the point of splitting them after the first weighing into possibly heavy or possibly light. Which of those 2 stacks has the 'fake coin' should evidently be too heavy if from the possibly heavy pile, or too light if from the possibly light pile.
That is to say, assuming that I'm left with 8 possibilities after using the scale for the first time, I know that in one set of 4, one of the coins may be lighter OR one of the coins from the other set may be heavier. One of these 2 conditions must be true for the results of the first 4/4/4 split to have been imbalanced, and the final determination of which coin is 'fake' will also determine whether it is heavy or light, depending on which stack it came from.
It's a slightly different approach than the other solution that was posted in the thread, but still believe it handles all outcomes.
Like any puzzle games, theres always an answers tab that we can check. Put the answers just like a result tab of a live report thread for people with low iq like me who cant even answer a single question after trying so hard.
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free. let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him. On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how?
REMEMBER/rules 1) there are 2 black balls in the bag and they are not changed 2) he does not have a trick ball anywhere - he chooses a ball from the bag 3) the ball he chooses is the one he is represented by 4)where he is
In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
Isn't this one all about division to circuits? The prisoners must decide to enforce an order on the boxes. Therefor, they decide to randomly send each person to a different box (dubbed "his" box), and the guy opens the box, and goes to the box belonging to the name inside it, according to the what they defined as that person's box. Basically, you enforce an order of circuits here, the chances of a circuit being 51 or greater are very close to ln2 (combinatorics for defining the permutation, some calculus to calculate the limit of the series needed to explain this number). Therefor, the chance of success would be (1 - ln2)*100 percent, somewhere around 30%. Not quite sure though.
Incorrect phrasing. The manager has 25 dollars in his pocket, his son has 2, and each of the guys has one (out of the 30). No dollar unaccounted for. The trick is that they payed 27, 25 are with the manager and 2 are with his son. No dollar vanished.
Each domino has to be on one black square and one white square. You eliminated 2 of the same color, now there are 32 of one color and 30 of the other, therefor, it will be impossible to put the last domino, as the only two squares left are of the same color, therefor not next to eachother.
25 and 26 were instant, 16 took me a bit too long. Will get back to the rest later.
On April 10 2011 18:45 0x64 wrote: In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
He pulls one out of the bag, and does something with it so that no one can retrieve it. Maybe he swallows it, and now it's in his digestive system. Maybe he throws it off the cliff before anyone gets a good look at it. The point is, everyone knows he just picked a ball, but they didn't see the color, and now they have no way of directly seeing it's color because the ball is gone. Lost. Irretrievable FOREVER!!!!!
So the only way they can determine the color of the ball is through logic. They open the bag and look at the remaining ball. It's black? Oh, then the one that he picked first MUST have been white!!!
Instead of thinking about the merging and reincarnating separately, think of the two combined as one process. So what are our options? 1.) Two high templar merge, and when that archon dies, it will reincarnate as a high templar. 2.) Two dark templar merge, and when that archon dies, it will reincarnate as a high templar. 3.) One of each templar merge, and when that archon dies, it will reincarnate as a dark templar.
If we perform "Action 1" on our initial group of templar, it is equivalent to removing a high templar from the group. Our new set is the initial set, minus one high templar. Cool.
If we perform "Action 2" on our initial group of templar, then our new set is the initial one, plus one high templar, minus two dark templar. Also cool.
If we perform "Action 3" on our initial group of templar, it has the same effect as Action 1. We just lose a high templar. Our new set is the initial set, again minus one high templar. SUPER COOL.
Notice the effect of each action in terms of parity. All three of these do not change the parity of the number of dark templar, and all of them DO change the parity of the number of high templar. So...
If we have an even number of dark templar to begin with, we will have an even number of dark templar at the end. Namely, we will have zero dark templar at the end, and our last templar is a high templar.
If we have an odd number of dark templar to begin with, we will have an odd number of dark templar at the end. Namely, we will have one dark templar. He will be the last templar, just sitting their chilling. Perhaps we could write a screenplay about him, much like "The Last Samurai," but in this case he'd be "The Last Templar." Of course, he would still be played by Tom Cruise.
Label and divide the coins in to three groups: ABCD EFGH IJKL.
Start off by balancing ABCD and EFGH Three possibilities, and I will address the simplest ones first: 1) ABCD = EFGH, find the fake coin in IJKL in two tries, you should know how to solve this, weigh ABC against IJK, and then I against J if it doesn't balance. 2) ABCD < EFGH, this is the same as the next one, just switch labels
3) ABCD > EFGH Weigh ABCEF against DIJKL, remember IJKL are now real coins and we have three possiblities:
1. ABCEF=DIJKL, we have now cleared A B C E F D by balancing the scale, we know one between G and H is fake and it weighs less than a real coin, balance G against H and find the lighter one.
2. ABCEF>DIJKL, D is clear because it was in the heavier group in step two, E and F are clear because they could only be lighter if they are fake. One of the A B C is fake and the fake coin is heavier, balance A against B and find the fake coin.
3. ABCEF<DIJKL, A B C are clear since they could only be heavier if they were fake. D E F are suspects here, either D is too heavy or E / F is too light. Balance E against F and find the fake coin.
Edit: just found out that everhate's solution from last page is much less convoluted, i somehow managed to axe that idea when brainstorming and came up with this ugly beast instead >_<.
You ask any of the daughters "is she younger than her (pointing to the two remaining sisters)" Based off of this information, you pick the daughter that is indicated to be younger.
There are 3 possible variations: 1. you asked the youngest 2. you asked the middle and 3. you asked the oldest
1. She will answer that the older daughter is the youngest (which is a lie) and you will pick the oldest daughter 2. You will not pick the middle daughter because she is the one that you asked the question to, and that is all that matters 3. She will answer that the youngest daughter is the youngest, and you will pick her.
Wow, beautiful!
That logic doesnt work because the younger sister could lie and tell u the middle one is the youngest. U could also ask the older sister if the middle one is younger and than her and she will answer u yes.
U could ask do u think lying to someone is ok? If u pick the younger one she will tell u no(lie) If u pick the older one she will tell u no If u pick the middle one she will answer yes because she lies sometimes and tell the truth other times
U then know if she answers yes that she is the middle sister and if she answers no u marrie her
No, yours doesn't work and the one you quoted works fine. The middle sister's answer is essentially random so you cannot ask a question that results in any possibility of marrying the person you asked. As a result, you have to ask a yes/no answer where yes = marrying one of the other two and no = marrying the other; if you get the middle sister the answer conveniently doesn't matter as both are eligible and thus the randomness is a nonfactor.
So, all you have to do is find a question which produces one answer for [NOT middle] and one answer for [middle] regardless of whether it's asked to the eldest or the youngest. The age question is such a question, because
-From the perspective of the eldest sister, she is the eldest and the liar is the youngest (which is the truth), therefore the middle sister is older than the liar.
-From the perspective of the youngest sister, she is the eldest and the truthteller is the youngest (the opposite of reality, since the youngest sister always lies), therefore she will also say that the middle sister is older than the liar. Remember that the youngest is compelled to lie and cannot choose not to, so her answer will be consistent.
So you ask sister A "is sister B older than sister C?".
-If sister A is the middle sister, either B or C are fine so the answer doesn't matter. -If sister A is the eldest and B is the middle sister, she will answer "yes" so you should pick C (the younger = liar) -If sister A is the eldest and B is the youngest, she will answer "no" so you should pick B (the younger = liar) -If sister A is the youngest and B is the middle, she will answer "yes" so you should pick C (the older = truthteller) -If sister A is the youngest and B is the eldest, she will answer "no" so you should pick B (the older = truthteller)
So we end up with
-If sister A answers "yes", pick C. Otherwise, pick B. This works for all permutations of A,B,C.
Edit: Didn't see a solution for 18 in the thread, so in case anyone wants it (it's incredibly simple algebra):
He could pick a black one, but then get them to check what was left in the bag. Since a black ball is left they would "know" he drew a white one and he would go free. And if things didn't go to plan he would at least expose that he was being cheated and might get a "fair" shot at avoiding justice. Not sure what any of this would have to do with where he is. I guess he could "accidentally" drop his black ball off the cliff to get them to check the bag.
If we say that the dark templar is 1 and the high templar is 0, we have a xor operation (see http://en.wikipedia.org/wiki/Exclusive_or )As we know, the XOR operation is commutative (the order doesn't matter) so we can merge in the order we want. Let's say we have M dark templar and N high templar : Let's merge all the dark templars together : if M is even, we get one high templar, otherwise we get one dark templar. Let's merge all the high templars together : we get an high templar. (ie 0 XOR 0 XOR 0 XOR 0... =0)
Result : If M is even, we get an high templar. If M is odd, we get an dark templar.
On April 10 2011 13:03 stepover12 wrote: 26. suppose you have a chess board with 2 opposite corners cut out as in picture there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
Your chessboard is sideways. In proper chess diagrams, The board is shown as if the white and black pieces were on the bottom and top of the board, respectively. An empty board would look like this: Just a nitpick. + Show Spoiler [solution] +
Answer would be no. each domino piece would, by necessity, cover 1 black square and 1 white square. Because there are 32 white squares and 30 black squares, once you've placed 30 dominos, the 31st would have to cover 2 white squares, which is not possible.
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
On April 11 2011 01:05 Tintti wrote: A puzzle of mine:
"Let's imagine a universe where there is a finite number of things people may know. Let's now assume that Albert want's to know everything there is to know. What is the last thing he shall know?"
Let's imagine a universe where there is a finite number of things people may know.
Is that a limit on the number of things an individual may know, or a limit on the number of things that are knowable?
The latter, although I'm not really sure how they actually differ :D
The wording I used in this riddle in my opinion suggests the correct answer more easily than the original one, which was something like "What's the last thing one may ever know?" I reckoned that as the riddlers here would anyway ask for clarifications on this rather obscure question I'd might as well post that in a more precise form.
although I'm not really sure how they actually differ
The difference between you being able to only hold 7 facts in your mind versus there only being 7 facts in the whole universe. If it was a limit on his mind + Show Spoiler +
then the last thing he'd know is that he couldn't know everything.
One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
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Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
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In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
- sell 3-liter jug - empty one 8-liter jug into pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 3+4 with 4 liters of water each - fill one 8-liter jug with the 8 liter water from pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 1+2 with exactly 4 liters - attack Terran with 4-pool
Treat high templar as the number +1 and dark templar as -1. Multiply all of them templar together to get the type of templar left at the end. So odd # of dark templar --> dark templar at end. Even # of dark templar --> high templar at end.
On April 11 2011 06:34 MusicalPulse wrote: One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
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i assume you can't just take 8 and split into 4-4 using equal water levels as that would make the problem trivial that said, it's still trivial, i never understood how these are riddles. i just go into it and start pouring willy nilly, knowing that it'll solve itself at some point.. here's the solution i came up with + Show Spoiler +
- sell 3-liter jug - empty one 8-liter jug into pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 3+4 with 4 liters of water each - fill one 8-liter jug with the 8 liter water from pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 1+2 with exactly 4 liters
Fill 3, put in 8 (3), fill 3, put in 8 (6). Fill 3, put in 8 (3), fill 3, put in 8 (6). Pour 8a (6) in 8b (6) -> 8a (4), 8b (8) Pour 8a into one of the infinite pools.
Repeat 3 times after emptying jugs.
this doesn't work btw
edit: re the stars question this immediately popped into my head
if you haven't seen her videos i encourage you to. if you're a nerd that is
- sell 3-liter jug - empty one 8-liter jug into pool 1 - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water in them - fill pools 3+4 with 4 liters of water each - fill one 8-liter jug with the 8 liter water from pool 1 in them - fill pools 1+2 with exactly 4 liters - fill the empty 8-liter jug from the full 8-liter jug until both have the same amount of water
Fill 3, put in 8 (3), fill 3, put in 8 (6). Fill 3, put in 8 (3), fill 3, put in 8 (6). Pour 8a (6) in 8b (6) -> 8a (4), 8b (8) Pour 8a into one of the infinite pools.
although I'm not really sure how they actually differ
The difference between you being able to only hold 7 facts in your mind versus there only being 7 facts in the whole universe. If it was a limit on his mind + Show Spoiler +
then the last thing he'd know is that he couldn't know everything.
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
To clarify yours - is the overlap between an odd number of squares all added together (for example, if I overlap them in such a way that they resemble a Venn Diagram, do the mergers between 1 and 2, 2 and 3, and 1 and 3 add to the merger of 1, 2, and 3 to form the area covered?
There are now 32 black squares and 30 white squares and each domino always covers two neighboring squares. The neighboring squares are always opposite in color. Due to this, the first 30 dominoes on the board must cover the 30 white squares and 30 black squares. You will then have one domino remaining. However there will be two black squares remaining. One domino only covers two opposite color squares so therefore it will not be able to cover the chess board.
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
They would just be stacked on top of each other, not adding to the initial area of 1. Because the overlap of multiple squares adds nothing to this area of one square, it is the minimum area of overlap. Unless I misunderstood your question (why do they HAVE to overlap, if you are asking about the net minimum area of overlap? why an odd number of squares, it seems irrelevant to my solution?).
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
They would just be stacked on top of each other, not adding to the initial area of 1. Because the overlap of multiple squares adds nothing to this area of one square, it is the minimum area of overlap. Unless I misunderstood your question (why do they HAVE to overlap, if you are asking about the net minimum area of overlap? why an odd number of squares, it seems irrelevant to my solution?).
I'm asking you to prove that there is no way to place the 1001 squares such that the area covered by an odd number of squares (see picture below) is less than 1. If you notice, if you overlap 2 squares on top of eachother, they cancel eachother's areas out.
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
To clarify yours - is the overlap between an odd number of squares all added together (for example, if I overlap them in such a way that they resemble a Venn Diagram, do the mergers between 1 and 2, 2 and 3, and 1 and 3 add to the merger of 1, 2, and 3 to form the area covered?
I'll draw a picture (mspaint ftw)
I made it circles instead of squares because it's easier to draw. The total area that would count would be the red, blue, yellow, (all covered by 1 circle) and black (covered by 3 circles) parts. The orange, green, and purple parts would not count, because they are covered by 2 circles each.
Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
Can you take the water out of the pools once you pour it in? Doesn't seem reasonable for "infinite" pools, not that infinite pools are reasonable (or in any way necessary for the problem?)
P.S. you can use [ code ] [ /code ] tags to monospace your answers
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
They would just be stacked on top of each other, not adding to the initial area of 1. Because the overlap of multiple squares adds nothing to this area of one square, it is the minimum area of overlap. Unless I misunderstood your question (why do they HAVE to overlap, if you are asking about the net minimum area of overlap? why an odd number of squares, it seems irrelevant to my solution?).
I'm asking you to prove that there is no way to place the 1001 squares such that the area covered by an odd number of squares (see picture below) is less than 1. If you notice, if you overlap 2 squares on top of eachother, they cancel eachother's areas out.
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
To clarify yours - is the overlap between an odd number of squares all added together (for example, if I overlap them in such a way that they resemble a Venn Diagram, do the mergers between 1 and 2, 2 and 3, and 1 and 3 add to the merger of 1, 2, and 3 to form the area covered?
I'll draw a picture (mspaint ftw)
I made it circles instead of squares because it's easier to draw. The total area that would count would be the red, blue, yellow, (all covered by 1 circle) and black (covered by 3 circles) parts. The orange, green, and purple parts would not count, because they are covered by 2 circles each.
Um so lets say I put 999 squares in a grid of 9 x 111, then in the top right corner of this grid I put one square that overlaps with the corner square by say .1 area, then one right on top of that one. So there are 3 squares overlapping an area of .1, 2 squares overlapping an area of .9 (which doesn't matter since they cancel their area out [wtf do you mean by that btw]), and the rest of the squares are not overlapping. Would that not defy the premise?
Um so lets say I put 999 squares in a grid of 9 x 111, then in the top right corner of this grid I put one square that overlaps with the corner square by say .1 area, then one right on top of that one. So there are 3 squares overlapping an area of .1, 2 squares overlapping an area of .9 (which doesn't matter since they cancel their area out [wtf do you mean by that btw]), and the rest of the squares are not overlapping. Would that not defy the premise?
The ones not overlapping would be a single layer, ie an odd number.
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
There is not a 50-50 chance that the guy you are shooting does not shoot you back if someone shot me I'd be dead certain to shoot that fucker back.
However If Black kills White then he has a 2/3 chance of dying. If he doesn't then he has a 0 chance of dying, Gray will shoot white next. If White is still alive, he will shoot Gray. If White is dead, Black will shoot Gray. If Black kills Gray then he has a 1 chance of dying. If he doesn't then he has a 0 chance of dying, Gray will shoot White next. If White is still alive, he will shoot Gray. If White is dead, Black will shoot Gray.
It's pretty obvious that Black wants Gray alive as long as possible, because Gray shoot White and White will kill Gray on his first shot, so it's more profitable for him to shoot White.
Um so lets say I put 999 squares in a grid of 9 x 111, then in the top right corner of this grid I put one square that overlaps with the corner square by say .1 area, then one right on top of that one. So there are 3 squares overlapping an area of .1, 2 squares overlapping an area of .9 (which doesn't matter since they cancel their area out [wtf do you mean by that btw]), and the rest of the squares are not overlapping. Would that not defy the premise?
The ones not overlapping would be a single layer, ie an odd number.
So why does a grid of squares that do not overlap at all not work? The area of overlap in this case is 0.
Can you take the water out of the pools once you pour it in? Doesn't seem reasonable for "infinite" pools, not that infinite pools are reasonable (or in any way necessary for the problem?)
P.S. you can use [ code ] [ /code ] tags to monospace your answers
Threads like these make me think that the starcraft community is the smartest community! =D
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?
Um so lets say I put 999 squares in a grid of 9 x 111, then in the top right corner of this grid I put one square that overlaps with the corner square by say .1 area, then one right on top of that one. So there are 3 squares overlapping an area of .1, 2 squares overlapping an area of .9 (which doesn't matter since they cancel their area out [wtf do you mean by that btw]), and the rest of the squares are not overlapping. Would that not defy the premise?
The ones not overlapping would be a single layer, ie an odd number.
So why does a grid of squares that do not overlap at all not work? The area of overlap in this case is 0.
It's an odd number of squares overlapping. 1 is an odd number, so it counts. Basically, a square sitting by itself still counts as an area of 1. (Single layer)
"A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?"
He has an affair with a woman on the 7th floor and goes there only when there is no one around. See, his wife has a phobia for rain, so whenever it rains she stays indoors and he has to come home ASAP to comfort her because he got rushed into the marriage for promises of money from her rich father that just doesn't seem to die at a young age and did not foresee the father living until he was 120. He is still waiting on that money, but his life is so miserable with the girl he married just for money, so he has an affair. When he is done banging the girl on the 7th floor he walks up the flight of stairs. He does this to make the sweat more pronounced on his clothes because he then claims that he just came back from the gym to minimize suspicion.
Um so lets say I put 999 squares in a grid of 9 x 111, then in the top right corner of this grid I put one square that overlaps with the corner square by say .1 area, then one right on top of that one. So there are 3 squares overlapping an area of .1, 2 squares overlapping an area of .9 (which doesn't matter since they cancel their area out [wtf do you mean by that btw]), and the rest of the squares are not overlapping. Would that not defy the premise?
The ones not overlapping would be a single layer, ie an odd number.
So why does a grid of squares that do not overlap at all not work? The area of overlap in this case is 0.
It's an odd number of squares overlapping. 1 is an odd number, so it counts. Basically, a square sitting by itself still counts as an area of 1. (Single layer)
1001:2 = 500 remainder 1. Because of this, there will always be 1 square that is either impeding on a stack of 2 making it 3 and therefore odd, or not overlapping with anything and therefore being 1 layer of its own, or both and in summation still being 1 unit of odd-layer space.
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
EDIT: Single layer counts as an odd number btw.
Oh ok, now I understand your question. Let me rephrase: Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
On April 11 2011 07:15 pikagrue wrote: Why my math thing no on front page?! T___T
You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
This has a really nice 1-2 line solution.
EDIT: Single layer counts as an odd number btw.
Oh ok, now I understand your question. Let me rephrase: Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
Edit: it's up on front page, #31
You got it! Your english skills are better than mine, lol
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
wrong, if he aims his first shot at mr. white, there's a chance that mr. white will be more inclined to shoot him first. if he aims at mr white he has a 1/3 chance of dying if he aims at mr gray he has a 2/3 chance of dying (as if he kills mr gray he will certainly die)
if he aims at nobody (shoots in the sky) mr. gray will shoot at mr. white due to the fact if he doesn't he will certainly die as he is the only one with a shot left, and if he misses mr. white will be more inclined to shoot mr. gray than mr. black due to the fact he tried to kill him.
i believe this amounts to a 0% chance of dying if he fires upwards.
On April 11 2011 08:22 Zeroes wrote: Threads like these make me think that the starcraft community is the smartest community! =D
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
i have a feeling this is written terribly and the original riddle is much more clever i can say "there are 3 coins on the table". this is true, and he will give me a coin. unfortunately there's nothing in the riddle that states which coin he will give me and how my statement is relevant at all. moreover it doesn't even state that the coin he gives you is one of the 3 that's on the table all in all, this needs a serious re-write
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
wrong, if he aims his first shot at mr. white, there's a chance that mr. white will be more inclined to shoot him first. if he aims at mr white he has a 1/3 chance of dying if he aims at mr gray he has a 2/3 chance of dying (as if he kills mr gray he will certainly die)
if he aims at nobody (shoots in the sky) mr. gray will shoot at mr. white due to the fact if he doesn't he will certainly die as he is the only one with a shot left, and if he misses mr. white will be more inclined to shoot mr. gray than mr. black due to the fact he tried to kill him.
i believe this amounts to a 0% chance of dying if he fires upwards.
Yep, I believe this is the right decision, although he still has a chance of dying if Mr. Gray misses
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
wrong, if he aims his first shot at mr. white, there's a chance that mr. white will be more inclined to shoot him first. if he aims at mr white he has a 1/3 chance of dying if he aims at mr gray he has a 2/3 chance of dying (as if he kills mr gray he will certainly die)
if he aims at nobody (shoots in the sky) mr. gray will shoot at mr. white due to the fact if he doesn't he will certainly die as he is the only one with a shot left, and if he misses mr. white will be more inclined to shoot mr. gray than mr. black due to the fact he tried to kill him.
i believe this amounts to a 0% chance of dying if he fires upwards.
Haha I didn't know he could should at nothing I was pondering over the implications of him wanting to shoot himself but still. White would not should Black just because he shot him first, Gray is gonna shoot white as well, so they are equally guilty and Gray is the bigger threat. Sorry.
Also that definitely does not amount to 0% chance of death because eventually Gray or White will die and Black will be the next one to shoot but only has 1/3 chance of shooting, and the next person will have a greater chance. Sorry bud, I think you're wrong on both ends here.
On April 11 2011 08:22 Zeroes wrote: Threads like these make me think that the starcraft community is the smartest community! =D
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
i have a feeling this is written terribly and the original riddle is much more clever i can say "there are 3 coins on the table". this is true, and he will give me a coin. unfortunately there's nothing in the riddle that states which coin he will give me and how my statement is relevant at all. moreover it doesn't even state that the coin he gives you is one of the 3 that's on the table all in all, this needs a serious re-write
There is an answer and the question doesn't need a re-write
On April 11 2011 08:22 Zeroes wrote: Threads like these make me think that the starcraft community is the smartest community! =D
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
i have a feeling this is written terribly and the original riddle is much more clever i can say "there are 3 coins on the table". this is true, and he will give me a coin. unfortunately there's nothing in the riddle that states which coin he will give me and how my statement is relevant at all. moreover it doesn't even state that the coin he gives you is one of the 3 that's on the table all in all, this needs a serious re-write
There is an answer and the question doesn't need a re-write
if you feel that way, then allow me to trivialize the problem by looking into the future: + Show Spoiler +
'you won't give me a non-gold coin'
true: you get a coin, and by asserting the truthfulness of the statement, that coin must be gold false: this would imply that he WILL give you a non-gold coin, but he cannot do that as per the rules of the game, thus only the 'true' interpretation works
added spoilers although i disagree
edit:
A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?
he made a bet with his buddy some time ago, and lost it. as a result, now he has to walk up 3 flights of stairs every day unless it is raining or he's with other people on the way home. the reason he does it on the way back from work as opposed to work is because otherwise he would get sweaty before work (did i mention he's hella fat) and be fired for not being professional.
On April 11 2011 08:50 pikagrue wrote: You have place 1001 unit squares on a coordinate plane. The squares can overlap (any number of squares can overlap in any fashion). Prove that the minimum amount of area where an odd number of squares overlap (amount of area covered by an odd number of squares) is equal to 1. The sides of the squares are parallel to the X and Y axes
Map each point (x,y) to the point in [0,1)^2 given by ({x},{y}). The image of each square is [0,1)^2, so each point in that is covered an odd number of times. When we consider the preimages of each point, at least one must be covered an odd number of times, solving the problem.
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
wrong, if he aims his first shot at mr. white, there's a chance that mr. white will be more inclined to shoot him first. if he aims at mr white he has a 1/3 chance of dying if he aims at mr gray he has a 2/3 chance of dying (as if he kills mr gray he will certainly die)
if he aims at nobody (shoots in the sky) mr. gray will shoot at mr. white due to the fact if he doesn't he will certainly die as he is the only one with a shot left, and if he misses mr. white will be more inclined to shoot mr. gray than mr. black due to the fact he tried to kill him.
i believe this amounts to a 0% chance of dying if he fires upwards.
Haha I didn't know he could should at nothing I was pondering over the implications of him wanting to shoot himself but still. White would not should Black just because he shot him first, Gray is gonna shoot white as well, so they are equally guilty and Gray is the bigger threat. Sorry.
Also that definitely does not amount to 0% chance of death because eventually Gray or White will die and Black will be the next one to shoot but only has 1/3 chance of shooting, and the next person will have a greater chance. Sorry bud, I think you're wrong on both ends here.
where is the error in this?
if gray shoots at white, white will shoot back at gray. there is no reason logically that he will shoot at black (this is how you are supposed to approach these riddles) .
there is nowhere in there that states that they will have no bias due to the attempt to kill them.
if gray shoots white, white will shoot gray. gray will shoot white because he is the biggest threat, and if he misses then the only likely scenario is that white shoots gray for an attempt at revenge.
On April 11 2011 08:35 Murderotica wrote: "A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?"
He has an affair with a woman on the 7th floor and goes there only when there is no one around. See, his wife has a phobia for rain, so whenever it rains she stays indoors and he has to come home ASAP to comfort her because he got rushed into the marriage for promises of money from her rich father that just doesn't seem to die at a young age and did not foresee the father living until he was 120. He is still waiting on that money, but his life is so miserable with the girl he married just for money, so he has an affair. When he is done banging the girl on the 7th floor he walks up the flight of stairs. He does this to make the sweat more pronounced on his clothes because he then claims that he just came back from the gym to minimize suspicion.
I don't know where I remember this, but your answer is wrong.
The man is short, so when he's alone on the elevator, he can only reach the 7th floor, otherwise he just asks when people are around, or uses his umbrella if its rainy.
On April 11 2011 08:35 Murderotica wrote: "A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?"
He has an affair with a woman on the 7th floor and goes there only when there is no one around. See, his wife has a phobia for rain, so whenever it rains she stays indoors and he has to come home ASAP to comfort her because he got rushed into the marriage for promises of money from her rich father that just doesn't seem to die at a young age and did not foresee the father living until he was 120. He is still waiting on that money, but his life is so miserable with the girl he married just for money, so he has an affair. When he is done banging the girl on the 7th floor he walks up the flight of stairs. He does this to make the sweat more pronounced on his clothes because he then claims that he just came back from the gym to minimize suspicion.
I don't know where I remember this, but your answer is wrong.
The man is short, so when he's alone on the elevator, he can only reach the 7th floor, otherwise he just asks when people are around, or uses his umbrella if its rainy.
On April 11 2011 08:35 Murderotica wrote: "A man who lives on the tenth floor takes the elevator down to the first floor every morning and goes to work. In the evening, when he comes back; on a rainy day, or if there are other people in the elevator, he goes to his floor directly. Otherwise, he goes to the seventh floor and walks up three flights of stairs to his apartment. Can you explain why?"
He has an affair with a woman on the 7th floor and goes there only when there is no one around. See, his wife has a phobia for rain, so whenever it rains she stays indoors and he has to come home ASAP to comfort her because he got rushed into the marriage for promises of money from her rich father that just doesn't seem to die at a young age and did not foresee the father living until he was 120. He is still waiting on that money, but his life is so miserable with the girl he married just for money, so he has an affair. When he is done banging the girl on the 7th floor he walks up the flight of stairs. He does this to make the sweat more pronounced on his clothes because he then claims that he just came back from the gym to minimize suspicion.
I don't know where I remember this, but your answer is wrong.
The man is short, so when he's alone on the elevator, he can only reach the 7th floor, otherwise he just asks when people are around, or uses his umbrella if its rainy.
your answer is correct as long as your story explains everything. lol.
I misinterpreted 8 as "two strings that take an hour to burn if you burn the first then burn the second once the first is done" so I was very confused for a while.
As to 13: most posters in this thread are missing the point of the problem. Yes, there is a binary expression for every natural number. The point is to PROVE that such an expression always exists (the proof was stated earlier in the thread)
On April 11 2011 07:45 Housemd wrote: Okay, I think I'll add one that isn't on the thread.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
he has to aim his first shot at mr white, because if he aims at grey and he hits him, mr white will kill him for sure
wrong, if he aims his first shot at mr. white, there's a chance that mr. white will be more inclined to shoot him first. if he aims at mr white he has a 1/3 chance of dying if he aims at mr gray he has a 2/3 chance of dying (as if he kills mr gray he will certainly die)
if he aims at nobody (shoots in the sky) mr. gray will shoot at mr. white due to the fact if he doesn't he will certainly die as he is the only one with a shot left, and if he misses mr. white will be more inclined to shoot mr. gray than mr. black due to the fact he tried to kill him.
i believe this amounts to a 0% chance of dying if he fires upwards.
Haha I didn't know he could should at nothing I was pondering over the implications of him wanting to shoot himself but still. White would not should Black just because he shot him first, Gray is gonna shoot white as well, so they are equally guilty and Gray is the bigger threat. Sorry.
Also that definitely does not amount to 0% chance of death because eventually Gray or White will die and Black will be the next one to shoot but only has 1/3 chance of shooting, and the next person will have a greater chance. Sorry bud, I think you're wrong on both ends here.
where is the error in this?
if gray shoots at white, white will shoot back at gray. there is no reason logically that he will shoot at black (this is how you are supposed to approach these riddles) .
there is nowhere in there that states that they will have no bias due to the attempt to kill them.
if gray shoots white, white will shoot gray. gray will shoot white because he is the biggest threat, and if he misses then the only likely scenario is that white shoots gray for an attempt at revenge.
if black shoots grey and hits, he loses. because white kills him if black shoots white and hits, he will normally still lose, because grey has a better accuracy than him. if black shoots grey and misses, grey still has to shoot white, because miss or hit white will choose to attack grey. if black shoots white and misses, grey should shoot white aswell because he has a better win rate vs black 1v1, white will then shoot grey because he is more likely to survive another shot from black.
in either miss situation grey and white still need to shoot each other as they are the greater threats, so it doesnt matter who he shoots at if he misses, but on the off chance he actually hits he wants to hit white, giving him a chance to still win.
he should aim at white, but its true, his chances are better if he misses first shot, so if he can he should shoot noone but its NOT because he doesnt want to anger anyone
you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
1. Fill the 3-liter bottle and pour it into the empty 5-liter bottle. 2. Fill the 3-liter bottle again, and pour enough to fill the 5-liter bottle. (This leaves exactly 1 liter in the 3-liter bottle.) 3. Empty the 5-liter bottle; pour the remaining 1 liter from the 3-liter bottle into
the 5-liter bottle. 4. Fill the 3-liter bottle and pour it into the 5-liter bottle. The 5-liter bottle
Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third
bag, etc. on the scale. If each coin were authentic, the total weight should be 55 grams. If the counterfeit coin is in bag #1, the total weight will be 54.9 grams. If the counterfeit coin is in bag #2, the total weight will be 54.8 grams. If the counterfeit coin is in bag #3, the total weight will be 54.7 grams. etc...
Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag(s) contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 4 coins from the third
bag, and 8 coins from the fourth bag. If all coins were authentic, the total weight should be 15 grams. Bag 1 only - 14.9 grams Bag 1,2 only - 14.7 grams Bag 1,3 only - 14.5 grams Bag 1,4 only - 14.1 grams Bag 1,2,3 only - 14.3 grams Bag 1,2,3,4 only - 13.5 grams Bag 2 only - 14.8 grams Bag 2,3 only - 14.4 grams Bag 2,4 only - 14.0 grams Bag 2,3,4 only - 13.6 grams Bag 3 only - 14.6 grams Bag 3,4 only - 13.8 grams Bag 4 only - 14.2 grams etc.
You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
Start with both hourglasses running. When the 7 minute hourglass runs out, invert it.
4 minutes later the 11 minute hourglass will run out. At this point 11 minutes will
have elapsed and if you turn over the 7 minute hourglass now, it will be 4 minutes
until it runs out, exactly 15 minutes. 11 + 4 = 15. Note: FUCK YEAH, INFINITE APM FTW. Time to go play zerg and burrow roach micro like a
A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
In t hours, the minute hand completes t revolutions. In the same amount of time, the
hour hand completes t/12 revolutions. The first time the minute hand and the hour hand overlap, the minute hand would have
completed 1 lap more than the hour hand. So we have t = t/12 + 1. This implies that
the first overlap happens after t = 12/11 hours (~1:05 pm).
Suppose a rectangle can be (in some way) entirely covered by 25 circular disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Let's take the simplest example of this form. Let's take a square (just another kind
of rectangle ^^) with a diagonal of 2 (each side of the square is the square root of
2). This circle is fully covered by one circle with radius of 1. Can this rectangle
be covered by 4 times as many circles of half the radius? First break up the square
into 4 quarters. This forms 4 more squares with a diagnoal of 1. Each of our 4
circles with a radius of 1/2 will cover each of the 4 squares. So the answer to the
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
On the 100th day, all 100 blue-eyed people will leave. If there is only one blue-eyed
person, he will see that there is no other blue-eyed person and then will leave the
island, knowing he is the one being referred to. If there are 2 blue-eyed people,
they will see the other and know if they are the only blue-eyed person that they will
leave that night. If they do not, then both of them leave on the 2nd night. This
process repeats until on the 100th day, all 100 blue-eyed people leave. The process
is difficult to understand intuitively and it relies on common knowledge ordered
Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
Take any eight of the nine coins, and load the scale up with four coins on either
side. If the two sides are equal, then the remaining coin is the fake. If the two sides are not equal, then the remaining coin is a real coin and the fake
one is on one side or the other of the scale. Now unload at the same time a single
coin from each of the scales. If the scales balance, the bad coin is one of the two
which you just withdrew. If the scales remain unbalanced, the fake is still on the
scales. As you remove good coins, you can add them to the "good coin pile" which
began with the first isolated coin. Once you have found the two coins which when
removed balance the scales, or if they are the final two and the scales are still
unbalanced, you take one of those two and weigh it against a known good coin. If they
balanced on the second loading of the scales, or if they don't, you have now with
only two loadings of the scales CORRECTLY IDENTIFIED THE FAKE COIN.
When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
First let's identify remainders of 30 that are prime and non-prime. Non-Prime - 2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28 Prime - 7,11,13,17,19,23,29 We should then strive to identify why a prime number greater than 32 divided by 30
(with a non-1 remainder) will NOT result in a remainder that is non-prime. First, we should remove any remainder that has a mulitple of 2, as 32/30, 34/30,
36/30, 38/30, could not be a prime number as it has a common factor of 2. Non-Prime - 3,5,9,15,21,25,27 Prime - 7,11,13,17,19,23,29 Second, we should remove any remainder that has a multiple of 5, as 35/30, 40/30,
45/30, could not be a prime number as it has a common factor of 5. Non-Prime - 3,9,15,21,27 Prime - 7,11,13,17,19,23,29 Third, we should remove any remainder that has a multiple of 3, as 33/30, 36/30,
39/30, could not be a prime number as it has a common factor of 3. Non-Prime - Prime - 7,11,13,17,19,23,29 Thus, because we are using prime numbers as our dividend, the remainder must always
Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
The viziers can use a binary code where each blue hat = 0 and each red hat = 1. The
prisoner in the back of the line adds up all the values of the hats he sees before
him and if the sum is even he says "blue" and if the sum is odd he says "red". This
prisoner has a 50/50 chance of having the hat color that he said, but each subsequent
prisoner can calculate his own color by adding up the hats in front (and behind after
hearing the answers) and comparing it to the initial answer given by the prisoner in
Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
The answer is no. Let us choose one side of the 13-gon, and consider the
parallelogram it belongs to (it is clear that there are not two such parallelograms).
The opposite side of this parallelogram is also a side of a second parallelogram.
This second parallelogram has another side parallel to the first, and we can continue
this "chain" of parallelograms until we arrive at a side of the 13-gon. This side is
therefore parallel to the side with which we started and since a convex polygon
cannot have three mutually parallel sides it is parallel to no other side of the
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Assuming the host behavior is that the car is placed randomly behind any door, and
the host must open a door revealing a goat regardless of the player's initial choice,
and if two doors are available the host chooses randomly. It is advantageous to
Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
Every natural number can be expressed as a sum of distinct powers of 2. The
expression is unique if written out in the form 2^x + 2^x+1 + 2^x+2, etc. regardless
of order. But if order is disregarded other forms of expression a sum of distinct
Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
The answer is yes if the collection of 100 distinct integers is consecutive. As you
could start with the largest number in the set, and set up the recursion A(n+1) = n -
7, such that you fill a set of 15 numbers that have a difference of seven between
terms in the set. If the collection of 100 distinct integers is for example multiples
of 3s (3,6,9,12,15,18,21,etc.) then no pair of numbers subtracted will be divisible
A room has 100 boxes labeled 1 thru 100. The names of 100 prisoners have been placed in these boxes by the warden. The prisoners shall visit the room one by one. Each prisoner is allowed to inspect the contents of at most 50 boxes, one after the other and leave the room with no communication with other prisoners. If the prisoner discovers his own name in the boxes he inspects, he is released. The prisoners are allowed to collude before hand and devise a strategy to maximize the chances of releasing each and every prisoner. What is their strategy?
The prisoners must agree on a random labeling of the boxes by their own names. When
admitted to the room, each prisoner inspects his own box (that is, the box with which
his own name has been associated). He then looks into the box belonging to the name
he just found, and then into the box belonging to the name he found in the second
box, etc. until he either finds his own name, or has opened 50 boxes. P.S. Here's how that ~30% probability is calculated. Let k > n and count the permutations having a cycle C of length exactly k. There are
(2n k) ways to pick the entries in C, (k-1)! ways to order them, and (2n-k)! ways to
permute the rest; the product of these numbers is (2n)!/k. Since at most one k-cycle
can exist in a given permutation, the probability that there is one is eactly 1/k. It follows that the probability that there is no long cycle is 1 - 1/(n+1) - 1/(n+2) - ... - 1/(2n) = 1 - H(2n) + H(n) where H(n) is the sum of the
reciprocals of the first n postivie integers, aproximately ln n. Thus our probability
is about 1 - ln 2n + ln n = 1 - ln , and in fact is always a bit larger. For n = 50
we get that the prisoners survive with probability ~31%.
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
Ask princess A "Is princess B older than princess C?" If princess A is the middle
princess, it doesn't matter which of the other two we choose. If princess A is the
eldest, we marry the one she indicates is younger. If princess A is the youngest, we
want to marry the elder of the other two, which means marrying the one she says is
younger. So if the answer is yes, we always marry princess C, and if it's no, we
A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Let x be the number of boats/men. There are x^2 people in total. 10 lifeboats sank
which each have x men; however, we saved 10 men in each of the remaining boats 10(x-
10). So this brings our expression to x^2 - 10x + 10(x-10) or x^2 - 10x + 10x - 100
or x^2 - 100. Since x^2 is the number of people we started with and x^2 - 100 is the
total number of survivors, we know that we have lost 100 people.
10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order * a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer how do they split the money and how many pirates die?
This problem can be solved by working backwards. Let's assume all but pirates 9 and
10 have been thrown overboard. Pirate 9 proposes all 100 gold coins for himself and 0
for Pirate 10. Since when he votes, he is at least 50% of the vote, he gets all the
money. If there are 3 pirates left (8, 9, and 10) 8 knows that 9 will offer 10 in the next
round; therefore, 8 has to offer Pirate 10 1 coin in this round to make Pirate 10
vote with him, and get his allocation thorugh. Therefore when only three are left the
allocation is Pirate 8: 99 Pirate 9: 0 Pirate 10: 1 If four pirates remain (7, 8, 9, and 10), 7 can offer 1 to pirate 9 to avoid being
thrown overboard. He cannot offer the same deal to pirate 10 as pirate 10 would just
as well get the gold from pirate 8, so would eagerly kill off pirate 7. Ultimtely this cycle of common knowledge occurs until: Pirate 1: 96 Pirate 2: 0 Pirate 3: 1 Pirate 4: 0 Pirate 5: 1 Pirate 6: 0 Pirate 7: 1 Pirate 8: 0 Pirate 9: 1 Pirate 10: 0
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer's plan was to drink a weak poison prior to the meeting with the king,
and then he would drink the pharmacist's strong poison, which would neutralize the
weak poison. As his own poison he would bring water, which will have no effect on
him, but the pharmacist who would drink the water, and then his poison would surely
die. When the pharmacist figured out this plan, he decided to bring water as well. So
the treasurer who drank poison earlier, drank the pharmacist's water, then his own
water, and died of the poison he drank before. The pharmacist would drink only water,
so nothing will happen to him. And because both of them brought the king water, he
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
The team nominates a leader. The leader is the only person who will announce that
everyone has visited the switch room. All the prisoners (except for the leader) will
flip the first switch up at their first opportunity, and again on the second
opportunity. If the first switch is already up, or they have already flipped the
first switch up two times, they will then flip the second switch. Only the leader may
flip the first switch down, if the first switch is already down, then the leader will
flip the second switch. The leader remembers how many times he has flipped the first
switch down. Once the leader has flipped the first switch down 44 times, he announces
that all have visited the room. It does not matter how many times a prisoner has
visited the room, in which order the prisoners were sent or even if the first switch
was initially up. Once the leader has flipped the switch down 44 times then the
leader knows everyone has visited the room. If the switch was initially down, then
all 22 prisoners will flip the switch up twice. If the switch was initially up, then
there will be one prisoner who only flips the switch up once and the rest will flip
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
This one was painful to prove, and I don't know how to represent it mathematically.
But in the smallest network of nodes that satisfy this (4), there are unique paths.
If you add an additional node, if it is the endpoint node, it does not change the
uniqueness of the paths you chose earlier. If the node is not the endpoint node, it
must have at least one path in and at least three paths out, and may reorder the
nodal path, but does not change the uniqueness. In proving that in the smallest 'n'
mode model n+1 nodes does not change the existence of a unique path, all models with
n modes that have connections to three other nodes are proven to have at least 2
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
Arbitrarily label the coins A-L First weighing: Left A,B,C,D Right E,F,G,H If they balance, the counterfeit is in I,J,K,L. If the left is heavier, the counterfeit coin is one of A,B,C,D and it is heavier or
the counterfeit coin is one of E,F,G,H and it is lighter If the right is heavier, the counterfeit coin is one of A,B,C,D and it is lighter or
the counterfeit coint is one of E,F,G,H and it is heavier Second weighing: Case 1: Left I,J,K Right A,B,C (if known good) If they balance, coin L is counterfeit. If the left is heavier, counterfeit coin is one of I,J,K and it is heavier If the right is heavier, counterfeit coin is one of I,J,K and it is lighter Case 2: Left A,B,C,E Right D,I,J,K If they balance, counterfeit coin is one of F,G,H and it is lighter If the left is heavier, counterfeit coin is one of A,B,C and it is heavier If the right is heavier, counterfeit coin is D and it is heavier or counterfeit coin
is E and it is lighter Third weighing: If counterfeit coin is known, but not whether it is heavy or light, compare the coin
with any of the others. If counterfeit coin is X and heavy or Y and light, compare X with a good coin. If X
is heavier then X is the counterfeit, else it is Y. If counterfeit coin is heavy and one of 3 coins (X,Y,Z) Compare X with Y. If X is heavier, then X is the coin. If Y is heavier, then Y is the
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
No. 1 and No. 2 go across: 2 minutes No. 2 returns with the flashlight: 2 minutes No. 3 and No. 4 go across: 10 minutes No. 1 returns with the flashlight: 1 minute No. 1 and No. 2 go across: 2 minutes 2 + 2 + 10 + 1 + 2 = 17 minutes
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
The equation 9x$3 = $27 is misleading. Here is an accounting of the $30 over time. Starting Time Man 1: $10 Man 2: $10 Man 3: $10 Manager: $0 Son: $0 $10+$10+$10+$0+$0 = $30 After Giving the Manager the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $30 Son: $0 $0+$0+$0+$30+$0 = $30 After Giving the Son the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $25 Son: $5 $0+$0+$0+$25+$5 = $30 After the Son takes $2 and Gives the Men Each $1 Man 1: $1 Man 2: $1 Man 3: $1 Manager: $25 Son: $2 $1+$1+$1+$25+$2 = $30 There is no missing dollar.
suppose you have a chess board with 2 opposite corners cut. there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
Since two diagonally opposite squares are the same color, it leaves 30 squares of a
color and 32 of another. Since a domino only covers two squares of opposite colors,
only 15 dominos at most can be fitted on the board.
In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
If two high templar merge, we are -1 high templar. If two dark templar merge, we are +1 high templar -2 dark templar. If one of each templar merge, we are -1 high templar. If we have an even number of dark templar to begin with, we will have an even number
of dark templar at the end. So our last templar will be a high templar. If we have an
odd number of dark templar to begin with, we will end with one dark templar at the
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
Let's label the jugs A, B, and C such that the first 8-liter jug is Jug A, the second
8-liter jug is Jug B, and the 3-liter jug is Jug C. Let's also label the pools pool
1, 2, 3, and 4. I was able to get a solution in 24 steps. A->C C->1 A->C A->2 C->A B->C C->A B->C C->A C->3 B->C A->C C->B A->C C->B A->C A->4 C->B C->1 B->C C->3 B->C C->4 B->2 And now all the jugs are empty and each pool has 4 liters of water.
Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
The vertices of an n-pointed star are the vertices of a regular n-gon, numbered 0
through n-1 in clockwise order. The star is determined by choosing a vertex m and
drawing the line segments from 0 to m, from m to 2m, from 2m to 3m, and (n-1)m to 0,
where all numbers are reduced modulo m. In order for the figure to satisfy our
conditions, m must be relatively prime to n and not equal to 1 or m-1. There are 400
positive numbers below 1000 that are relatively prime to 1000. Since the same star
results from choosing the first edge to go from 0 to k as when it goes from 0 to n-k,
In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
Since the bullet travels faster than the speed of sound (1116.44 fps at sea level) he
will feel the pain of a foot thoroughly ruined before he hears the shot.
Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
If all the squares are stacked up, then the area is one. If there is one even stack
and one odd stack, then the area is one. If there are any more than one odd stack,
then the area is greater than one. It is impossible to have no odd stacks. Note that a stack can be a series of overlapping squares or squares that overlap perfectly. But a stack will have a minimum area of one.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at
Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is
just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr.
Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr.
Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White,
Mr. Black has a better chance of winning than before.
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
"You will give me neither the copper or the silver coin." If it's true, you get the gold coin. If it's false, it breaks the conditions that you get no coin when lying.
And now to pretend that no new logic problems will pop up so I don't waste my entire life here.
you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
1. Fill the 3-liter bottle and pour it into the empty 5-liter bottle. 2. Fill the 3-liter bottle again, and pour enough to fill the 5-liter bottle. (This leaves exactly 1 liter in the 3-liter bottle.) 3. Empty the 5-liter bottle; pour the remaining 1 liter from the 3-liter bottle into
the 5-liter bottle. 4. Fill the 3-liter bottle and pour it into the 5-liter bottle. The 5-liter bottle
Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third
bag, etc. on the scale. If each coin were authentic, the total weight should be 55 grams. If the counterfeit coin is in bag #1, the total weight will be 54.9 grams. If the counterfeit coin is in bag #2, the total weight will be 54.8 grams. If the counterfeit coin is in bag #3, the total weight will be 54.7 grams. etc...
Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag(s) contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 4 coins from the third
bag, and 8 coins from the fourth bag. If all coins were authentic, the total weight should be 15 grams. Bag 1 only - 14.9 grams Bag 1,2 only - 14.7 grams Bag 1,3 only - 14.5 grams Bag 1,4 only - 14.1 grams Bag 1,2,3 only - 14.3 grams Bag 1,2,3,4 only - 13.5 grams Bag 2 only - 14.8 grams Bag 2,3 only - 14.4 grams Bag 2,4 only - 14.0 grams Bag 2,3,4 only - 13.6 grams Bag 3 only - 14.6 grams Bag 3,4 only - 13.8 grams Bag 4 only - 14.2 grams etc.
You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
Start with both hourglasses running. When the 7 minute hourglass runs out, invert it.
4 minutes later the 11 minute hourglass will run out. At this point 11 minutes will
have elapsed and if you turn over the 7 minute hourglass now, it will be 4 minutes
until it runs out, exactly 15 minutes. 11 + 4 = 15. Note: FUCK YEAH, INFINITE APM FTW. Time to go play zerg and burrow roach micro like a
A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
In t hours, the minute hand completes t revolutions. In the same amount of time, the
hour hand completes t/12 revolutions. The first time the minute hand and the hour hand overlap, the minute hand would have
completed 1 lap more than the hour hand. So we have t = t/12 + 1. This implies that
the first overlap happens after t = 12/11 hours (~1:05 pm).
Suppose a rectangle can be (in some way) entirely covered by 25 circular disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Let's take the simplest example of this form. Let's take a square (just another kind
of rectangle ^^) with a diagonal of 2 (each side of the square is the square root of
2). This circle is fully covered by one circle with radius of 1. Can this rectangle
be covered by 4 times as many circles of half the radius? First break up the square
into 4 quarters. This forms 4 more squares with a diagnoal of 1. Each of our 4
circles with a radius of 1/2 will cover each of the 4 squares. So the answer to the
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
On the 100th day, all 100 blue-eyed people will leave. If there is only one blue-eyed
person, he will see that there is no other blue-eyed person and then will leave the
island, knowing he is the one being referred to. If there are 2 blue-eyed people,
they will see the other and know if they are the only blue-eyed person that they will
leave that night. If they do not, then both of them leave on the 2nd night. This
process repeats until on the 100th day, all 100 blue-eyed people leave. The process
is difficult to understand intuitively and it relies on common knowledge ordered
Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
Take any eight of the nine coins, and load the scale up with four coins on either
side. If the two sides are equal, then the remaining coin is the fake. If the two sides are not equal, then the remaining coin is a real coin and the fake
one is on one side or the other of the scale. Now unload at the same time a single
coin from each of the scales. If the scales balance, the bad coin is one of the two
which you just withdrew. If the scales remain unbalanced, the fake is still on the
scales. As you remove good coins, you can add them to the "good coin pile" which
began with the first isolated coin. Once you have found the two coins which when
removed balance the scales, or if they are the final two and the scales are still
unbalanced, you take one of those two and weigh it against a known good coin. If they
balanced on the second loading of the scales, or if they don't, you have now with
only two loadings of the scales CORRECTLY IDENTIFIED THE FAKE COIN.
When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
First let's identify remainders of 30 that are prime and non-prime. Non-Prime - 2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28 Prime - 7,11,13,17,19,23,29 We should then strive to identify why a prime number greater than 32 divided by 30
(with a non-1 remainder) will NOT result in a remainder that is non-prime. First, we should remove any remainder that has a mulitple of 2, as 32/30, 34/30,
36/30, 38/30, could not be a prime number as it has a common factor of 2. Non-Prime - 3,5,9,15,21,25,27 Prime - 7,11,13,17,19,23,29 Second, we should remove any remainder that has a multiple of 5, as 35/30, 40/30,
45/30, could not be a prime number as it has a common factor of 5. Non-Prime - 3,9,15,21,27 Prime - 7,11,13,17,19,23,29 Third, we should remove any remainder that has a multiple of 3, as 33/30, 36/30,
39/30, could not be a prime number as it has a common factor of 3. Non-Prime - Prime - 7,11,13,17,19,23,29 Thus, because we are using prime numbers as our dividend, the remainder must always
Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
The viziers can use a binary code where each blue hat = 0 and each red hat = 1. The
prisoner in the back of the line adds up all the values of the hats he sees before
him and if the sum is even he says "blue" and if the sum is odd he says "red". This
prisoner has a 50/50 chance of having the hat color that he said, but each subsequent
prisoner can calculate his own color by adding up the hats in front (and behind after
hearing the answers) and comparing it to the initial answer given by the prisoner in
Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
The answer is no. Let us choose one side of the 13-gon, and consider the
parallelogram it belongs to (it is clear that there are not two such parallelograms).
The opposite side of this parallelogram is also a side of a second parallelogram.
This second parallelogram has another side parallel to the first, and we can continue
this "chain" of parallelograms until we arrive at a side of the 13-gon. This side is
therefore parallel to the side with which we started and since a convex polygon
cannot have three mutually parallel sides it is parallel to no other side of the
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Assuming the host behavior is that the car is placed randomly behind any door, and
the host must open a door revealing a goat regardless of the player's initial choice,
and if two doors are available the host chooses randomly. It is advantageous to
Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
Every natural number can be expressed as a sum of distinct powers of 2. The
expression is unique if written out in the form 2^x + 2^x+1 + 2^x+2, etc. regardless
of order. But if order is disregarded other forms of expression a sum of distinct
Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
The answer is yes if the collection of 100 distinct integers is consecutive. As you
could start with the largest number in the set, and set up the recursion A(n+1) = n -
7, such that you fill a set of 15 numbers that have a difference of seven between
terms in the set. If the collection of 100 distinct integers is for example multiples
of 3s (3,6,9,12,15,18,21,etc.) then no pair of numbers subtracted will be divisible
A room has 100 boxes labeled 1 thru 100. The names of 100 prisoners have been placed in these boxes by the warden. The prisoners shall visit the room one by one. Each prisoner is allowed to inspect the contents of at most 50 boxes, one after the other and leave the room with no communication with other prisoners. If the prisoner discovers his own name in the boxes he inspects, he is released. The prisoners are allowed to collude before hand and devise a strategy to maximize the chances of releasing each and every prisoner. What is their strategy?
The prisoners must agree on a random labeling of the boxes by their own names. When
admitted to the room, each prisoner inspects his own box (that is, the box with which
his own name has been associated). He then looks into the box belonging to the name
he just found, and then into the box belonging to the name he found in the second
box, etc. until he either finds his own name, or has opened 50 boxes. P.S. Here's how that ~30% probability is calculated. Let k > n and count the permutations having a cycle C of length exactly k. There are
(2n k) ways to pick the entries in C, (k-1)! ways to order them, and (2n-k)! ways to
permute the rest; the product of these numbers is (2n)!/k. Since at most one k-cycle
can exist in a given permutation, the probability that there is one is eactly 1/k. It follows that the probability that there is no long cycle is 1 - 1/(n+1) - 1/(n+2) - ... - 1/(2n) = 1 - H(2n) + H(n) where H(n) is the sum of the
reciprocals of the first n postivie integers, aproximately ln n. Thus our probability
is about 1 - ln 2n + ln n = 1 - ln , and in fact is always a bit larger. For n = 50
we get that the prisoners survive with probability ~31%.
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
Ask princess A "Is princess B older than princess C?" If princess A is the middle
princess, it doesn't matter which of the other two we choose. If princess A is the
eldest, we marry the one she indicates is younger. If princess A is the youngest, we
want to marry the elder of the other two, which means marrying the one she says is
younger. So if the answer is yes, we always marry princess C, and if it's no, we
A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Let x be the number of boats/men. There are x^2 people in total. 10 lifeboats sank
which each have x men; however, we saved 10 men in each of the remaining boats 10(x-
10). So this brings our expression to x^2 - 10x + 10(x-10) or x^2 - 10x + 10x - 100
or x^2 - 100. Since x^2 is the number of people we started with and x^2 - 100 is the
total number of survivors, we know that we have lost 100 people.
10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order * a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer how do they split the money and how many pirates die?
This problem can be solved by working backwards. Let's assume all but pirates 9 and
10 have been thrown overboard. Pirate 9 proposes all 100 gold coins for himself and 0
for Pirate 10. Since when he votes, he is at least 50% of the vote, he gets all the
money. If there are 3 pirates left (8, 9, and 10) 8 knows that 9 will offer 10 in the next
round; therefore, 8 has to offer Pirate 10 1 coin in this round to make Pirate 10
vote with him, and get his allocation thorugh. Therefore when only three are left the
allocation is Pirate 8: 99 Pirate 9: 0 Pirate 10: 1 If four pirates remain (7, 8, 9, and 10), 7 can offer 1 to pirate 9 to avoid being
thrown overboard. He cannot offer the same deal to pirate 10 as pirate 10 would just
as well get the gold from pirate 8, so would eagerly kill off pirate 7. Ultimtely this cycle of common knowledge occurs until: Pirate 1: 96 Pirate 2: 0 Pirate 3: 1 Pirate 4: 0 Pirate 5: 1 Pirate 6: 0 Pirate 7: 1 Pirate 8: 0 Pirate 9: 1 Pirate 10: 0
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer's plan was to drink a weak poison prior to the meeting with the king,
and then he would drink the pharmacist's strong poison, which would neutralize the
weak poison. As his own poison he would bring water, which will have no effect on
him, but the pharmacist who would drink the water, and then his poison would surely
die. When the pharmacist figured out this plan, he decided to bring water as well. So
the treasurer who drank poison earlier, drank the pharmacist's water, then his own
water, and died of the poison he drank before. The pharmacist would drink only water,
so nothing will happen to him. And because both of them brought the king water, he
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
The team nominates a leader. The leader is the only person who will announce that
everyone has visited the switch room. All the prisoners (except for the leader) will
flip the first switch up at their first opportunity, and again on the second
opportunity. If the first switch is already up, or they have already flipped the
first switch up two times, they will then flip the second switch. Only the leader may
flip the first switch down, if the first switch is already down, then the leader will
flip the second switch. The leader remembers how many times he has flipped the first
switch down. Once the leader has flipped the first switch down 44 times, he announces
that all have visited the room. It does not matter how many times a prisoner has
visited the room, in which order the prisoners were sent or even if the first switch
was initially up. Once the leader has flipped the switch down 44 times then the
leader knows everyone has visited the room. If the switch was initially down, then
all 22 prisoners will flip the switch up twice. If the switch was initially up, then
there will be one prisoner who only flips the switch up once and the rest will flip
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
This one was painful to prove, and I don't know how to represent it mathematically.
But in the smallest network of nodes that satisfy this (4), there are unique paths.
If you add an additional node, if it is the endpoint node, it does not change the
uniqueness of the paths you chose earlier. If the node is not the endpoint node, it
must have at least one path in and at least three paths out, and may reorder the
nodal path, but does not change the uniqueness. In proving that in the smallest 'n'
mode model n+1 nodes does not change the existence of a unique path, all models with
n modes that have connections to three other nodes are proven to have at least 2
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
Arbitrarily label the coins A-L First weighing: Left A,B,C,D Right E,F,G,H If they balance, the counterfeit is in I,J,K,L. If the left is heavier, the counterfeit coin is one of A,B,C,D and it is heavier or
the counterfeit coin is one of E,F,G,H and it is lighter If the right is heavier, the counterfeit coin is one of A,B,C,D and it is lighter or
the counterfeit coint is one of E,F,G,H and it is heavier Second weighing: Case 1: Left I,J,K Right A,B,C (if known good) If they balance, coin L is counterfeit. If the left is heavier, counterfeit coin is one of I,J,K and it is heavier If the right is heavier, counterfeit coin is one of I,J,K and it is lighter Case 2: Left A,B,C,E Right D,I,J,K If they balance, counterfeit coin is one of F,G,H and it is lighter If the left is heavier, counterfeit coin is one of A,B,C and it is heavier If the right is heavier, counterfeit coin is D and it is heavier or counterfeit coin
is E and it is lighter Third weighing: If counterfeit coin is known, but not whether it is heavy or light, compare the coin
with any of the others. If counterfeit coin is X and heavy or Y and light, compare X with a good coin. If X
is heavier then X is the counterfeit, else it is Y. If counterfeit coin is heavy and one of 3 coins (X,Y,Z) Compare X with Y. If X is heavier, then X is the coin. If Y is heavier, then Y is the
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
No. 1 and No. 2 go across: 2 minutes No. 2 returns with the flashlight: 2 minutes No. 3 and No. 4 go across: 10 minutes No. 1 returns with the flashlight: 1 minute No. 1 and No. 2 go across: 2 minutes 2 + 2 + 10 + 1 + 2 = 17 minutes
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
The equation 9x$3 = $27 is misleading. Here is an accounting of the $30 over time. Starting Time Man 1: $10 Man 2: $10 Man 3: $10 Manager: $0 Son: $0 $10+$10+$10+$0+$0 = $30 After Giving the Manager the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $30 Son: $0 $0+$0+$0+$30+$0 = $30 After Giving the Son the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $25 Son: $5 $0+$0+$0+$25+$5 = $30 After the Son takes $2 and Gives the Men Each $1 Man 1: $1 Man 2: $1 Man 3: $1 Manager: $25 Son: $2 $1+$1+$1+$25+$2 = $30 There is no missing dollar.
suppose you have a chess board with 2 opposite corners cut. there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
Since two diagonally opposite squares are the same color, it leaves 30 squares of a
color and 32 of another. Since a domino only covers two squares of opposite colors,
only 15 dominos at most can be fitted on the board.
In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
If two high templar merge, we are -1 high templar. If two dark templar merge, we are +1 high templar -2 dark templar. If one of each templar merge, we are -1 high templar. If we have an even number of dark templar to begin with, we will have an even number
of dark templar at the end. So our last templar will be a high templar. If we have an
odd number of dark templar to begin with, we will end with one dark templar at the
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
Let's label the jugs A, B, and C such that the first 8-liter jug is Jug A, the second
8-liter jug is Jug B, and the 3-liter jug is Jug C. Let's also label the pools pool
1, 2, 3, and 4. I was able to get a solution in 24 steps. A->C C->1 A->C A->2 C->A B->C C->A B->C C->A C->3 B->C A->C C->B A->C C->B A->C A->4 C->B C->1 B->C C->3 B->C C->4 B->2 And now all the jugs are empty and each pool has 4 liters of water.
Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
The vertices of an n-pointed star are the vertices of a regular n-gon, numbered 0
through n-1 in clockwise order. The star is determined by choosing a vertex m and
drawing the line segments from 0 to m, from m to 2m, from 2m to 3m, and (n-1)m to 0,
where all numbers are reduced modulo m. In order for the figure to satisfy our
conditions, m must be relatively prime to n and not equal to 1 or m-1. There are 400
positive numbers below 1000 that are relatively prime to 1000. Since the same star
results from choosing the first edge to go from 0 to k as when it goes from 0 to n-k,
In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
Since the bullet travels faster than the speed of sound (1116.44 fps at sea level) he
will feel the pain of a foot thoroughly ruined before he hears the shot.
Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
If all the squares are stacked up, then the area is one. If there is one even stack
and one odd stack, then the area is one. If there are any more than one odd stack,
then the area is greater than one. It is impossible to have no odd stacks. Note that a stack can be a series of overlapping squares or squares that overlap perfectly. But a stack will have a minimum area of one.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at
Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is
just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr.
Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr.
Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White,
Mr. Black has a better chance of winning than before.
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
"You will give me neither the copper or the silver coin." If it's true, you get the gold coin. If it's false, it breaks the conditions that you get no coin when lying.
And now to pretend that no new logic problems will pop up so I don't waste my entire life here.
Amazing! Check back once in a while I'll have new ones
For problem 32, should we assume gray and white are perfectly rational (will always choose a target as such to maximize their survival possibilities)? Also, is black allowed to purposefully miss?
First, if black kills gray, he dies to white. Simple. So the strategy of shooting at gray is infinitely inferior to purposefully missing, since purposefully missing guarantee's gray's turn comes whereas shooting at gray only gives a 2/3 chance of gray's turn coming. So black must either shoot at white or purposefully miss.
Let's say white's turn somehow comes around. If he shoots gray, there is a 2/3 chance he survives. If he shoots black, there is a 1/3 chance he survives. (This can be seen trivially by examining the possibilities). So white WILL shoot gray and gray will die if his turn comes.
So gray must shoot white and kill him or he will certainly die. If gray gets a chance to shoot at white, there is a (2/3) chance he kills him. Then its a duel with him and black with black getting the first shot. If black misses then gray hits, it's (2/3)(2/3). Or if gray misses once, it's (2/3)(1/3)(2/3)(2/3) he wins. And so on up to infinite misses from gray. Factoring out one of the (2/3) we can express the probability gray survives if he shoots at white (remember that if gray misses, he dies) as 4/9 times the sum from n = 1 to infinity of 2^n / 3^(2n + 1). This equals 8/21 chance of gray's survival.
If gray misses white (1/3 of the time) then white has a 2/3 chance of survival (shooting at gray). So white has a 2/9 chance of survival.
So, since the probabilities of survival equal one, these are the probabilities of survival if black shoots at white and misses or if he wastes his turn:
white: (14/63) gray: (24/63) black: (25/63)
That's right, if black wastes his turn or misses white, his probability of survival becomes the HIGHEST of the three!
So we must consider the probabilities if black shoots at white.
The probability is P(black kills white) * P(black kills gray when gray is allowed to shoot first) + P(black misses) * 25/63
Gray's chance of missing is 1/3. So if gray misses then black kills, it's (1/3)(1/3). If gray then black misses then gray misses then black kills, it's (1/3)(2/3)(1/3)(1/3). Then the next term in the sum is (1/3)(2/3)(1/3)(2/3)(1/3)(1/3).
Factoring out 1/3 and considering the possibilities of black killing on his nth shot, taking the sum from 1 to infinity:
1/3 * sum from n = 1 to infinity of 2^(n-1) / 3^(2n - 1)
That equals 1/7.
So, black's chance of survival if he shoots at white:
First consider how he would fair in 1v1 fights vGrey 3/7 (1st) 1/7 (2nd) vWhite 1/3 (1st) 0 (2nd)
black shoots grey H(1/3)grey dies, white shoots black 0 M(2/3)grey shoots white H(2/3)white dies, black shoots grey 3/7 (4/21) M(1/3)white shoots grey, grey dies, black shoots white 1/3 (2/27) (4/21+2/27)=26.46%
black shoots white H(1/3)white dies, grey shoots black 1/7 (1/21) M(2/3)grey shoots white H(2/3)white dies, black shoots grey 3/7 (4/21) M(1/3)white shoots grey, grey dies, black shoots white 1/3 (2/27) (1/21+4/21+2/27)=31.22%
black shoots noone grey shoots white H(2/3)white dies, black shoots grey 3/7 (2/7) M(1/3)white shoots grey 1/3 (1/9) (2/7+1/9)=39.68%
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
This one was painful to prove, and I don't know how to represent it mathematically.
But in the smallest network of nodes that satisfy this (4), there are unique paths.
If you add an additional node, if it is the endpoint node, it does not change the
uniqueness of the paths you chose earlier. If the node is not the endpoint node, it
must have at least one path in and at least three paths out, and may reorder the
nodal path, but does not change the uniqueness. In proving that in the smallest 'n'
mode model n+1 nodes does not change the existence of a unique path, all models with
n modes that have connections to three other nodes are proven to have at least 2
unique routes.
There are many, many things wrong with this solution, but first let me point out that the number of nodes has to be even for 3-regular graph (a network in which each node is connected to three other nodes). Your method uses induction which inherently assumes the number of nodes can be odd.
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
This one was painful to prove, and I don't know how to represent it mathematically.
But in the smallest network of nodes that satisfy this (4), there are unique paths.
If you add an additional node, if it is the endpoint node, it does not change the
uniqueness of the paths you chose earlier. If the node is not the endpoint node, it
must have at least one path in and at least three paths out, and may reorder the
nodal path, but does not change the uniqueness. In proving that in the smallest 'n'
mode model n+1 nodes does not change the existence of a unique path, all models with
n modes that have connections to three other nodes are proven to have at least 2
unique routes.
There are many, many things wrong with this solution, but first let me point out that the number of nodes has to be even for 3-regular graph (a network in which each node is connected to three other nodes). Your method uses induction which inherently assumes the number of nodes can be odd.
nice try though ;p
Even isn't the only restriction though, right?
The smallest 3-regular graph has 4 nodes.
I think the key to this problem is to first determine possible numbers of nodes that the graph can have
The first line intersects 0 lines. Then, the next line intersects 1, the next intersects 2...
Every Nth line intersects (N-1) lines at unique intersection points. So for a system of N lines, we take the arithmetic sum 0 + 1 + 2 + 3 + ... + N-1, which by the formula for arithmetic series equals N(N-1)/2 total unique intersections. When N = 10, the intersections number 45. (10 * 9 / 2)
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
This one was painful to prove, and I don't know how to represent it mathematically.
But in the smallest network of nodes that satisfy this (4), there are unique paths.
If you add an additional node, if it is the endpoint node, it does not change the
uniqueness of the paths you chose earlier. If the node is not the endpoint node, it
must have at least one path in and at least three paths out, and may reorder the
nodal path, but does not change the uniqueness. In proving that in the smallest 'n'
mode model n+1 nodes does not change the existence of a unique path, all models with
n modes that have connections to three other nodes are proven to have at least 2
unique routes.
There are many, many things wrong with this solution, but first let me point out that the number of nodes has to be even for 3-regular graph (a network in which each node is connected to three other nodes). Your method uses induction which inherently assumes the number of nodes can be odd.
nice try though ;p
Even isn't the only restriction though, right?
The smallest 3-regular graph has 4 nodes.
I think the key to this problem is to first determine possible numbers of nodes that the graph can have
A 3-regular graph can have any even number of nodes > 3. For example, construct such a 3-regular graph with x nodes by drawing a regular x-gon and pairing each vertex with another vertex to which it is not adjacent. This is not an important part of the solution at all though.
On April 11 2011 06:34 MusicalPulse wrote: One that's similar to the liters of water problem.
Here's what you have:
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
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Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
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In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
I don't understand what you mean in the Madadia one.... you can use kinematics to figure out at what angle the gun has to be pointed, but the first evidence should be the pain in his foot from being shot.... considering that the assassin is "hidden"
Everyone doing the Mr. White, Mr. Grey, Mr. Black thing is working way too hard. Once you realize that one of your options is to intentionally miss the first shot, you can easily see that your other two options are strictly worse without actually computing the percentages.
With math, the goal is to be as lazy as possible IMO.
On April 10 2011 06:32 tomnov wrote: [b]10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order
I think I figured this one out when I was bored at work today--but I think you could argue for 2 solutions (semantics!), and I'll post the logic first -- then the solutions in a simplified form.
Think about the situations backwards to figure out the criteria for satisfying pirates on earlier votes. Let's use the format #n for pirates, xG for their respective gold split, a simple (y/n) for if they vote yes on the proposed situation, and label pirates from captain P10 and the lowest rank pirate P1.
Starting with 2 pirates remaining: P2, 100G, y :: P1, 0G, n. 50% means this passes.
3 pirates remaining: P3, 99G, y :: P2, 0G, n :: P10, 1G, y. 66% means this passes; the logic indicates that pirate 1 cannot get a better deal at this point, so 1g is enough for pirate 3 to buy his vote to save his life; at this point, pirate 2 cannot be bribed but is mathematically irrelevant.
4 pirates remaining: P4, 99G, y :: P3 0G, n :: P2, 1G, y :: P1, 0G, n. 50% means this passes. Our incredibly logical P2 realizes that there's no way to get past the p3 situation, so can be bribed for his vote for only 1 gold.
5 pirates remaining: P5, 97G, y :: P4 0G, n :: P3, 1G, y :: P2, 0G, n :: P1 2G (or 1G depending), y. 60% means this passes. This may seem tricky, but it's the only logical way to justify the voting--and we see a pattern emerge. P5 can guarantee getting P3's and P1's votes for their respective bribes because they can't do any better in any later situations. The reason I say it might be able to be 1G or 2G for P1 is because the pirates that are getting paid out alternate, and so every iteration puts each set of pirates at risk of getting nothing--thus any situation where the get 1G is better than them getting 0G, and logically they should accept any situation where the get 1G: adding 1G to their pay out just works to clinch the deal with the deal for the pirate deciding because none of the of the other pirates would be able to do better.
We can see based on this patterning that none of the pirates will be killed, and it all comes down to the captain's distribution.
It will either look like for the guaranteed ideal satisfaction of the even pirates: P10, 90G, y :: P9, 0G, n :: P8, 1G, y :: P7, 0G, n :: P6, 2G, y :: P5, 0G, n :: P4, 3G, y :: P3, 0G, n :: P2, 4G, y :: P1, 0G, n.
or it will just be 96,0,1,0,1,0,1,0,1,0 if we're accounting for pirates' desire to minimize risk.
I don't understand what you mean in the Madadia one.... you can use kinematics to figure out at what angle the gun has to be pointed, but the first evidence should be the pain in his foot from being shot.... considering that the assassin is "hidden"
4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
Let's rephrase the problem. Given a 3-regular graph that has a Hamiltonian circuit, we want to show it has to have at least one more.
First some definitions:
Let a T-coloring of the graph be defined as a set of three disjoint classes of edges (which we will call T-classes) such that every edge in the graph belongs to one of the three classes and also such that every vertex is connected to one edge from each of the three classes. A T-coloring shall be unordered.
Let a S-subset be the union of simple cycles (A simple cycle is a cycle with no repeated vertices) such that each cycle contains an even number of edges and such that every vertex in the graph belongs to one of the cycles. Let n(S) be the number of cycles in S. When n(S)=1, we get a Hamiltonian cycle. For each edge in N, let its coefficient in X(S) be 1 if that edge belongs in S and 0 otherwise. X(S) shall be calculated mod 2.
Now onto the proof. We will prove that the sum of X(H) for all Hamiltonian circuits in N = 0 (mod 2). If there was only one Hamiltonian circuit, then that value will obviously not be 0.
For every T-coloring, we can associate 3 different S-subsets to it. This is what I mean by associate: take two T-classes and let their union form a S-subset. The third T-class is comprises of the edges that do not belong in that S-subset. It is easy to see that X(S_1)+X(S_2)+X(S_3) for the three subsets associated with the particular T-coloring = 0 (mod 2), as each edge occurs in two of the subsets. Call this result (1).
For every S-subset of N, there are 2^(n(S)-1) different T-colorings that associate with it. That is because in each cycle of the S-subset, the two T-classes that form it will alternate, and so each cycle can be colored in two ways. Note it is 2^(n(S)-1) not 2^(n(S)) because the T-coloring is unordered.
Now, the sum of 2^(n(S)-1) * X(S) over every S of N is equivalent to summing (1) over all T-colorings which equals 0 (mod 2). 2^(n(S)-1) * X(S) is naturally 0 (mod 2) for all S for which n(S) >1. Thus it follows that sum{2^(n(S)-1) * X(S)} = 0 for when n(S) = 1, i.e. sum{X(H)} = 0. Q.E.D.
On April 11 2011 11:23 PJA wrote: Everyone doing the Mr. White, Mr. Grey, Mr. Black thing is working way too hard. Once you realize that one of your options is to intentionally miss the first shot, you can easily see that your other two options are strictly worse without actually computing the percentages.
Comparing shooting at grey vs shooting at white is pretty straight forward, but comparing those to not shooting anyone seems non trivial. Are you just waving your hands or can you actually explain? And does it work for any ordered probabilities?
Like if black was a 1% shot, grey was a 1.1% shot, and white was 100%.
4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
#2 knows. If 2 and three had the same color, then 1 would know and would shout out his (the opposite of what both of them have). but because 1 is silent, 2 and 3 are different. So 2 can look at three and say the opposite.
Trying to think of some that fall under a "math" category...
You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
I think that the 2 guy knows his hat is blue. Given he knows that #3 has a red hat, but #1 hasn't said his color hat (if he saw 2 red hats he would know his hat is blue), then its safe to assume he has a blue hat. #4 seems irrelevant to the riddle at all as he cannot make any statement due to lack of information.
4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
#2 knows. If 2 and three had the same color, then 1 would know and would shout out his (the opposite of what both of them have). but because 1 is silent, 2 and 3 are different. So 2 can look at three and say the opposite.
Trying to think of some that fall under a "math" category...
You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
You add 1 teaspoon of pure tea. You take out 1/11th tsp of tea and 10/11ths of coffee. You initially put in 11/11th of tea so you have 10/11ths of tea left over. It's equal.
On April 11 2011 12:24 gyth wrote: graph theory.+ Show Spoiler +
Given a 3-regular graph that has a Hamiltonian circuit, we want to show it has to have at least one more...
...sum{X(H)} = 0. Q.E.D.
This is a classic result in graph theory.
What was the result? Did that mean it did or didn't have at least one more?
If there was only one H, sum{X(H)} would not equal 0. So there has to be another one at least. In fact, there has to be two more at least since it is not possible to have sum{X(H)} equal to 0 mod 2 with two H's either.
On April 11 2011 12:22 hacklebeast wrote: harder: + Show Spoiler +
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
On April 11 2011 12:36 Murderotica wrote: Harder math: + Show Spoiler +
You add 1 teaspoon of pure tea. You take out 1/11th tsp of tea and 10/11ths of coffee. You initially put in 11/11th of tea so you have 10/11ths of tea left over. It's equal.
Well, you take 1 teaspoon of pure tea and put it into the coffee cup. You then take a contaminated mix that is roughly 10 parts coffee to 1 part tea from the coffee cup and put that into the tea cup. There is clearly more tea in the coffee cup than there is coffee in the tea cup.
Problem ?
This is the easiest thing in the world lol.. you seemed halfway there Murderotica.. unless I'm missing something?
I don't understand what you mean in the Madadia one.... you can use kinematics to figure out at what angle the gun has to be pointed, but the first evidence should be the pain in his foot from being shot.... considering that the assassin is "hidden"
4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
#2 knows. If 2 and three had the same color, then 1 would know and would shout out his (the opposite of what both of them have). but because 1 is silent, 2 and 3 are different. So 2 can look at three and say the opposite.
Trying to think of some that fall under a "math" category...
You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
Divide the pile into two smaller ones of 50. One pile will contain n heads, the other will contain 50-n heads. Flip all the coins in one pile, which will reverse it to being n heads in both.
My god, this one is so annoying. I hate working with it - but the solution is as follows: First Step: AT B:10C1T Second Step: AT B:100/11C, 10/11T TEASPOON: 1/11th T, 10/11ths C Third Step: A:100/11T,10/11C B:100/11C, 10/11T
On April 11 2011 12:22 hacklebeast wrote: harder: + Show Spoiler +
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
On April 11 2011 12:36 Murderotica wrote: Harder math: + Show Spoiler +
You add 1 teaspoon of pure tea. You take out 1/11th tsp of tea and 10/11ths of coffee. You initially put in 11/11th of tea so you have 10/11ths of tea left over. It's equal.
Well, you take 1 teaspoon of pure tea and put it into the coffee cup. You then take a contaminated mix that is roughly 10 parts coffee to 1 part tea from the coffee cup and put that into the tea cup. There is clearly more tea in the coffee cup than there is coffee in the tea cup.
Problem ?
This is the easiest thing in the world lol.. you seemed halfway there Murderotica.. unless I'm missing something?
Lol let me explain it this way.
Let's say that there are 1100x in one cup and 1100y in the second. You take 1/10th of the 1100x and add it to the 1100y. This means your cups have: (1100-110)x and 1100y + 110x So, then you take 1/11th (remember there are now 11 teaspoons in the y cup so taking 1 out will be taking out 1/11th of the total inside of it) of the y/x mixed cup and add it to the x cup. So you have: (1100-110)x + (1100y + 110x)/11 and (1100y + 110x) * 10/11 This gives us:
At beginning 1) Start 7 minute and 11 Minute 2) As 7 minute end, re-flip it Total Time: 7 Minute 3) As 11 minute end, re-flip the 7 minute at THAT MOMENT Total Time: 11+4 = 15 Minute
This might be something everyone knows the answer to and might already have been posted(though I cant say I have seen it and I think i have read all of the 16 pages) But here goes.
You have to walk into 1 of 2 doors, where one will lead to certain instantenous death, and the other will lead to a life full of starcraft 2 and happy things. Each door is guarded by an all knowing all powerfull zergling(A crackling, if you will). One of the zerglings will always lie and the other will always tell the truth. You can ask One of them, 1 question. What do you ask to figure out what door to take? (You obviously do not know what zergling lies, and what door the zergling guards holds no relation to him lying or telling the truth.)
On April 11 2011 14:26 Earll wrote: This might be something everyone knows the answer to and might already have been posted(though I cant say I have seen it and I think i have read all of the 16 pages) But here goes.
You have to walk into 1 of 2 doors, where one will lead to certain instantenous death, and the other will lead to a life full of starcraft 2 and happy things. Each door is guarded by an all knowing all powerfull zergling(A crackling, if you will). One of the zerglings will always lie and the other will always tell the truth. You can ask One of them, 1 question. What do you ask to figure out what door to take? (You obviously do not know what zergling lies, and what door the zergling guards holds no relation to him lying or telling the truth.)
I heard this question b4... but not in the form of starcraft, you are awesome ^_^
On April 11 2011 14:26 Earll wrote: This might be something everyone knows the answer to and might already have been posted(though I cant say I have seen it and I think i have read all of the 16 pages) But here goes.
You have to walk into 1 of 2 doors, where one will lead to certain instantenous death, and the other will lead to a life full of starcraft 2 and happy things. Each door is guarded by an all knowing all powerfull zergling(A crackling, if you will). One of the zerglings will always lie and the other will always tell the truth. You can ask One of them, 1 question. What do you ask to figure out what door to take? (You obviously do not know what zergling lies, and what door the zergling guards holds no relation to him lying or telling the truth.)
Here's another one: 40. Suppose that 20 starcraft pros played in a tournament where each player competed against every other player exactly once. The result of each game is only win/lose, no draw. Is it always possible (regardless of results) to name the players 1,2,3,...,20 so that player 1 defeated player 2, player 2 defeated player 3,... player 19 defeated player 20? Give a proof or a counterexample.
you have a 5-liter jug and a 3-liter jug and a pool of water. How can you produce exactly 4 liters of water? (a classic one, appeared in a "die hard" movie lol)
1. Fill the 3-liter bottle and pour it into the empty 5-liter bottle. 2. Fill the 3-liter bottle again, and pour enough to fill the 5-liter bottle. (This leaves exactly 1 liter in the 3-liter bottle.) 3. Empty the 5-liter bottle; pour the remaining 1 liter from the 3-liter bottle into
the 5-liter bottle. 4. Fill the 3-liter bottle and pour it into the 5-liter bottle. The 5-liter bottle
Suppose we have 10 bags, each bag contains 10 coins. One of the bags contains counterfeit coins, the other 9 bags contain real coins. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 3 coins from the third
bag, etc. on the scale. If each coin were authentic, the total weight should be 55 grams. If the counterfeit coin is in bag #1, the total weight will be 54.9 grams. If the counterfeit coin is in bag #2, the total weight will be 54.8 grams. If the counterfeit coin is in bag #3, the total weight will be 54.7 grams. etc...
Suppose we have 4 bags, each bag contains 10 coins. Some of the bags contains counterfeit all coins, some contain all real coins. We don't know how many bags of counterfeit coins there are. Each counterfeit coin weighs 0.9 grams. Each real coin weighs 1.0 grams. If we have an accurate scale that give exact weight of whatever is placed on, could we determine which bag(s) contains the counterfeit coins with just _one_ weighing?
Place 1 coin from the first bag, 2 coins from the second bag, 4 coins from the third
bag, and 8 coins from the fourth bag. If all coins were authentic, the total weight should be 15 grams. Bag 1 only - 14.9 grams Bag 1,2 only - 14.7 grams Bag 1,3 only - 14.5 grams Bag 1,4 only - 14.1 grams Bag 1,2,3 only - 14.3 grams Bag 1,2,3,4 only - 13.5 grams Bag 2 only - 14.8 grams Bag 2,3 only - 14.4 grams Bag 2,4 only - 14.0 grams Bag 2,3,4 only - 13.6 grams Bag 3 only - 14.6 grams Bag 3,4 only - 13.8 grams Bag 4 only - 14.2 grams etc.
You have 2 hour-glasses, one measuring 7 minutes and the other 11 minutes. You want to boil an egg for exactly 15 minutes. Can you use the 2 hour-glasses to measure exactly 15 minutes? Note: your hands are so high APM it takes infinitely small amount of time to flip an hour glass.
Start with both hourglasses running. When the 7 minute hourglass runs out, invert it.
4 minutes later the 11 minute hourglass will run out. At this point 11 minutes will
have elapsed and if you turn over the 7 minute hourglass now, it will be 4 minutes
until it runs out, exactly 15 minutes. 11 + 4 = 15. Note: FUCK YEAH, INFINITE APM FTW. Time to go play zerg and burrow roach micro like a
A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide?
In t hours, the minute hand completes t revolutions. In the same amount of time, the
hour hand completes t/12 revolutions. The first time the minute hand and the hour hand overlap, the minute hand would have
completed 1 lap more than the hour hand. So we have t = t/12 + 1. This implies that
the first overlap happens after t = 12/11 hours (~1:05 pm).
Suppose a rectangle can be (in some way) entirely covered by 25 circular disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Note: overlaps allowed of course.
Let's take the simplest example of this form. Let's take a square (just another kind
of rectangle ^^) with a diagonal of 2 (each side of the square is the square root of
2). This circle is fully covered by one circle with radius of 1. Can this rectangle
be covered by 4 times as many circles of half the radius? First break up the square
into 4 quarters. This forms 4 more squares with a diagnoal of 1. Each of our 4
circles with a radius of 1/2 will cover each of the 4 squares. So the answer to the
A group of people with assorted eye colors live on an island. They are all perfect logicians -- if a conclusion can be logically deduced, they will do it instantly. No one knows the color of their eyes. Every night at midnight, a ferry stops at the island. Any islanders who have figured out the color of their own eyes then leave the island, and the rest stay. Everyone can see everyone else at all times and keeps a count of the number of people they see with each eye color (excluding themselves), but they cannot otherwise communicate. Everyone on the island knows all the rules in this paragraph.
On this island there are 100 blue-eyed people, 100 brown-eyed people, and the Guru (she happens to have green eyes). So any given blue-eyed person can see 100 people with brown eyes and 99 people with blue eyes (and one with green), but that does not tell him his own eye color; as far as he knows the totals could be 101 brown and 99 blue. Or 100 brown, 99 blue, and he could have red eyes.
The Guru is allowed to speak once (let's say at noon), on one day in all their endless years on the island. Standing before the islanders, she says the following:
"I can see someone who has blue eyes."
Who leaves the island, and on what night?
There are no mirrors or reflecting surfaces, nothing dumb. It is not a trick question, and the answer is logical. It doesn't depend on tricky wording or anyone lying or guessing, and it doesn't involve people doing something silly like creating a sign language or doing genetics. The Guru is not making eye contact with anyone in particular; she's simply saying "I count at least one blue-eyed person on this island who isn't me."
And lastly, the answer is not "no one leaves."
On the 100th day, all 100 blue-eyed people will leave. If there is only one blue-eyed
person, he will see that there is no other blue-eyed person and then will leave the
island, knowing he is the one being referred to. If there are 2 blue-eyed people,
they will see the other and know if they are the only blue-eyed person that they will
leave that night. If they do not, then both of them leave on the 2nd night. This
process repeats until on the 100th day, all 100 blue-eyed people leave. The process
is difficult to understand intuitively and it relies on common knowledge ordered
Suppose we have 9 coins that look the same and feel the same. But exactly one of them is counterfeit and it weighs less than a real coin. Can we identify the counterfeit coin among the 9 coins with just two weighings on an accurate balance scale?
Take any eight of the nine coins, and load the scale up with four coins on either
side. If the two sides are equal, then the remaining coin is the fake. If the two sides are not equal, then the remaining coin is a real coin and the fake
one is on one side or the other of the scale. Now unload at the same time a single
coin from each of the scales. If the scales balance, the bad coin is one of the two
which you just withdrew. If the scales remain unbalanced, the fake is still on the
scales. As you remove good coins, you can add them to the "good coin pile" which
began with the first isolated coin. Once you have found the two coins which when
removed balance the scales, or if they are the final two and the scales are still
unbalanced, you take one of those two and weigh it against a known good coin. If they
balanced on the second loading of the scales, or if they don't, you have now with
only two loadings of the scales CORRECTLY IDENTIFIED THE FAKE COIN.
When a prime number greater than 32 is divided by 30, you get a remainder R. If R is not equal to 1, must the remainder R be a prime number? Why or why not?
First let's identify remainders of 30 that are prime and non-prime. Non-Prime - 2,3,4,5,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28 Prime - 7,11,13,17,19,23,29 We should then strive to identify why a prime number greater than 32 divided by 30
(with a non-1 remainder) will NOT result in a remainder that is non-prime. First, we should remove any remainder that has a mulitple of 2, as 32/30, 34/30,
36/30, 38/30, could not be a prime number as it has a common factor of 2. Non-Prime - 3,5,9,15,21,25,27 Prime - 7,11,13,17,19,23,29 Second, we should remove any remainder that has a multiple of 5, as 35/30, 40/30,
45/30, could not be a prime number as it has a common factor of 5. Non-Prime - 3,9,15,21,27 Prime - 7,11,13,17,19,23,29 Third, we should remove any remainder that has a multiple of 3, as 33/30, 36/30,
39/30, could not be a prime number as it has a common factor of 3. Non-Prime - Prime - 7,11,13,17,19,23,29 Thus, because we are using prime numbers as our dividend, the remainder must always
Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)?
The viziers can use a binary code where each blue hat = 0 and each red hat = 1. The
prisoner in the back of the line adds up all the values of the hats he sees before
him and if the sum is even he says "blue" and if the sum is odd he says "red". This
prisoner has a 50/50 chance of having the hat color that he said, but each subsequent
prisoner can calculate his own color by adding up the hats in front (and behind after
hearing the answers) and comparing it to the initial answer given by the prisoner in
Can a convex 13-gon be tiled (partitioned) by parallelograms? (A 13-gon is a solid polygon of 13 sides. "Convex" means the straight line segment connecting any 2 points of the polygon lie inside the polygon. "Tile" meaning the overlaps between parallelograms can only happen at their edges.)
The answer is no. Let us choose one side of the 13-gon, and consider the
parallelogram it belongs to (it is clear that there are not two such parallelograms).
The opposite side of this parallelogram is also a side of a second parallelogram.
This second parallelogram has another side parallel to the first, and we can continue
this "chain" of parallelograms until we arrive at a side of the 13-gon. This side is
therefore parallel to the side with which we started and since a convex polygon
cannot have three mutually parallel sides it is parallel to no other side of the
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Assuming the host behavior is that the car is placed randomly behind any door, and
the host must open a door revealing a goat regardless of the player's initial choice,
and if two doors are available the host chooses randomly. It is advantageous to
Can every natural number (e.g.1,2,3,...) be expressed as a sum of distinct powers of 2 (e.g.1,2,4,8,...)? If so, is that expression unique (ignoring order of the terms in the sum)?
Every natural number can be expressed as a sum of distinct powers of 2. The
expression is unique if written out in the form 2^x + 2^x+1 + 2^x+2, etc. regardless
of order. But if order is disregarded other forms of expression a sum of distinct
Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
The answer is yes if the collection of 100 distinct integers is consecutive. As you
could start with the largest number in the set, and set up the recursion A(n+1) = n -
7, such that you fill a set of 15 numbers that have a difference of seven between
terms in the set. If the collection of 100 distinct integers is for example multiples
of 3s (3,6,9,12,15,18,21,etc.) then no pair of numbers subtracted will be divisible
A room has 100 boxes labeled 1 thru 100. The names of 100 prisoners have been placed in these boxes by the warden. The prisoners shall visit the room one by one. Each prisoner is allowed to inspect the contents of at most 50 boxes, one after the other and leave the room with no communication with other prisoners. If the prisoner discovers his own name in the boxes he inspects, he is released. The prisoners are allowed to collude before hand and devise a strategy to maximize the chances of releasing each and every prisoner. What is their strategy?
The prisoners must agree on a random labeling of the boxes by their own names. When
admitted to the room, each prisoner inspects his own box (that is, the box with which
his own name has been associated). He then looks into the box belonging to the name
he just found, and then into the box belonging to the name he found in the second
box, etc. until he either finds his own name, or has opened 50 boxes. P.S. Here's how that ~30% probability is calculated. Let k > n and count the permutations having a cycle C of length exactly k. There are
(2n k) ways to pick the entries in C, (k-1)! ways to order them, and (2n-k)! ways to
permute the rest; the product of these numbers is (2n)!/k. Since at most one k-cycle
can exist in a given permutation, the probability that there is one is eactly 1/k. It follows that the probability that there is no long cycle is 1 - 1/(n+1) - 1/(n+2) - ... - 1/(2n) = 1 - H(2n) + H(n) where H(n) is the sum of the
reciprocals of the first n postivie integers, aproximately ln n. Thus our probability
is about 1 - ln 2n + ln n = 1 - ln , and in fact is always a bit larger. For n = 50
we get that the prisoners survive with probability ~31%.
You are the most eligible bachelor in the kingdom, and as such the King has invited you to his castle so that you may choose one of his three daughters to marry. The eldest princess is honest and always tells the truth. The youngest princess is dishonest and always lies. The middle princess is mischievous and tells the truth sometimes and lies the rest of the time.
As you will be forever married to one of the princesses, you want to marry the eldest (truth-teller) or the youngest (liar) because at least you know where you stand with them.
The problem is that you cannot tell which sister is which just by their appearance, and the King will only grant you ONE yes or no question which you may only address to ONE of the sisters. What yes or no question can you ask which will ensure you do not marry the middle sister? Clarification: The answer you get wil ONLY be “yes” or “no” and you cannot ask a question that seeks a different answer or communicate with the daughters in any other way.
Ask princess A "Is princess B older than princess C?" If princess A is the middle
princess, it doesn't matter which of the other two we choose. If princess A is the
eldest, we marry the one she indicates is younger. If princess A is the youngest, we
want to marry the elder of the other two, which means marrying the one she says is
younger. So if the answer is yes, we always marry princess C, and if it's no, we
A ship had distributed the crew names on the many lifeboats onboard. Each lifeboat had equally many men, and there were exactly the same amount of men in each boat as there were boats in all.
During a storm the ship began to sink, and 10 lifeboats were destroyed by the waves with an unknown amount of men. The remaining crew pulled an additional 10 men into each of the remaining lifeboats.
How many drowned?
Let x be the number of boats/men. There are x^2 people in total. 10 lifeboats sank
which each have x men; however, we saved 10 men in each of the remaining boats 10(x-
10). So this brings our expression to x^2 - 10x + 10(x-10) or x^2 - 10x + 10x - 100
or x^2 - 100. Since x^2 is the number of people we started with and x^2 - 100 is the
total number of survivors, we know that we have lost 100 people.
10 pirates found a loot of 100 gold pieces, and decided to split it the following way: the captain offers how to split it, then they hold a vote and if at least half of them agree that is the split, else (more than half disagree) they kill him and the next in command tries, they vote again, and so on. the pirates want to stay alive, get the most gold, and kill the most of the other pirates in that order * a pirate will offer a split where he gets 0 gold if he knows that any other split will not get the votes and he will die * a pirate will not vote for a split if he knows he can get the same gold from the next pirate to offer how do they split the money and how many pirates die?
This problem can be solved by working backwards. Let's assume all but pirates 9 and
10 have been thrown overboard. Pirate 9 proposes all 100 gold coins for himself and 0
for Pirate 10. Since when he votes, he is at least 50% of the vote, he gets all the
money. If there are 3 pirates left (8, 9, and 10) 8 knows that 9 will offer 10 in the next
round; therefore, 8 has to offer Pirate 10 1 coin in this round to make Pirate 10
vote with him, and get his allocation thorugh. Therefore when only three are left the
allocation is Pirate 8: 99 Pirate 9: 0 Pirate 10: 1 If four pirates remain (7, 8, 9, and 10), 7 can offer 1 to pirate 9 to avoid being
thrown overboard. He cannot offer the same deal to pirate 10 as pirate 10 would just
as well get the gold from pirate 8, so would eagerly kill off pirate 7. Ultimtely this cycle of common knowledge occurs until: Pirate 1: 96 Pirate 2: 0 Pirate 3: 1 Pirate 4: 0 Pirate 5: 1 Pirate 6: 0 Pirate 7: 1 Pirate 8: 0 Pirate 9: 1 Pirate 10: 0
In a far away land, it was known that if you drank poison, the only way to save yourself is to drink a stronger poison, which neutralizes the weaker poison. The king that ruled the land wanted to make sure that he possessed the strongest poison in the kingdom, in order to ensure his survival, in any situation. So the king called the kingdom's pharmacist and the kingdom's treasurer, he gave each a week to make the strongest poison. Then, each would drink the other one's poison, then his own, and the one that will survive, will be the one that had the stronger poison. The pharmacist went straight to work, but the treasurer knew he had no chance, for the pharmacist was much more experienced in this field, so instead, he made up a plan to survive and make sure the pharmacist dies. On the last day the pharmacist suddenly realized that the treasurer would know he had no chance, so he must have a plan. After a little thought, the pharmacist realized what the treasurer's plan must be, and he concocted a counter plan, to make sure he survives and the treasurer dies. When the time came, the king summoned both of them. They drank the poisons as planned, and the treasurer died, the pharmacist survived, and the king didn't get what he wanted. What exactly happened there?
The treasurer's plan was to drink a weak poison prior to the meeting with the king,
and then he would drink the pharmacist's strong poison, which would neutralize the
weak poison. As his own poison he would bring water, which will have no effect on
him, but the pharmacist who would drink the water, and then his poison would surely
die. When the pharmacist figured out this plan, he decided to bring water as well. So
the treasurer who drank poison earlier, drank the pharmacist's water, then his own
water, and died of the poison he drank before. The pharmacist would drink only water,
so nothing will happen to him. And because both of them brought the king water, he
The warden meets with 23 new prisoners when they arrive. He tells them, "You may meet today and plan a strategy. But after today, you will be in isolated cells and will have no communication with one another.
"In the prison there is a switch room which contains two light switches labeled A and B, each of which can be in either the 'on' or the 'off' position. The switches are not connected to anything.
"After today, from time to time whenever I feel so inclined, I will select one prisoner at random and escort him to the switch room. This prisoner will select one of the two switches and reverse its position. He must move one, but only one of the switches. He can't move both but he can't move none either. Then he'll be led back to his cell."
"No one else will enter the switch room until I lead the next prisoner there, and he'll be instructed to do the same thing. I'm going to choose prisoners at random. I may choose the same guy three times in a row, or I may jump around and come back."
"But, given enough time, everyone will eventually visit the switch room as many times as everyone else. At any time any one of you may declare to me, 'We have all visited the switch room.'
"If it is true, then you will all be set free. If it is false, and somebody has not yet visited the switch room, you will be fed to the alligators."
*note - the only difference from Scenario B, the original position of the 2 switches are known.
Assuming that:
A) There is no restriction on the amount of time the prisoners could take before sending the notice to the warden that everyone has been to the switch room at least once.
B) There is no restriction on the number of time each prisoner can visit the switch room
C) The warden will not attempt any foul moves, such as intentionally not bringing a certain prisoner to the switch room forever.
The team nominates a leader. The leader is the only person who will announce that
everyone has visited the switch room. All the prisoners (except for the leader) will
flip the first switch up at their first opportunity, and again on the second
opportunity. If the first switch is already up, or they have already flipped the
first switch up two times, they will then flip the second switch. Only the leader may
flip the first switch down, if the first switch is already down, then the leader will
flip the second switch. The leader remembers how many times he has flipped the first
switch down. Once the leader has flipped the first switch down 44 times, he announces
that all have visited the room. It does not matter how many times a prisoner has
visited the room, in which order the prisoners were sent or even if the first switch
was initially up. Once the leader has flipped the switch down 44 times then the
leader knows everyone has visited the room. If the switch was initially down, then
all 22 prisoners will flip the switch up twice. If the switch was initially up, then
there will be one prisoner who only flips the switch up once and the rest will flip
A young zergling hero from Zerus wants to explore the land his race has conquered. To do this, he wants to visit every zerg planet exactly once using nydus canals and return to his home planet. Every one of these planets is connected to exactly three other planets by nydus canals. He has already planned a route but does not like it for some reason. Is there another route he can take? If so prove its existence. *Note the new route cannot just be the reverse of the original route.
My solution to this problem was fundamentally wrong. Luckily, someone much smarter than me was able to figure it out. As such, we should consider him a gentleman and a scholar.
On April 11 2011 12:15 LastPrime wrote: I don't think anyone is going to get #22 so I'll post the solution here. It is quite long and "mathy" so be warned.
Let's rephrase the problem. Given a 3-regular graph that has a Hamiltonian circuit, we want to show it has to have at least one more.
First some definitions:
Let a T-coloring of the graph be defined as a set of three disjoint classes of edges (which we will call T-classes) such that every edge in the graph belongs to one of the three classes and also such that every vertex is connected to one edge from each of the three classes. A T-coloring shall be unordered.
Let a S-subset be the union of simple cycles (A simple cycle is a cycle with no repeated vertices) such that each cycle contains an even number of edges and such that every vertex in the graph belongs to one of the cycles. Let n(S) be the number of cycles in S. When n(S)=1, we get a Hamiltonian cycle. For each edge in N, let its coefficient in X(S) be 1 if that edge belongs in S and 0 otherwise. X(S) shall be calculated mod 2.
Now onto the proof. We will prove that the sum of X(H) for all Hamiltonian circuits in N = 0 (mod 2). If there was only one Hamiltonian circuit, then that value will obviously not be 0.
For every T-coloring, we can associate 3 different S-subsets to it. This is what I mean by associate: take two T-classes and let their union form a S-subset. The third T-class is comprises of the edges that do not belong in that S-subset. It is easy to see that X(S_1)+X(S_2)+X(S_3) for the three subsets associated with the particular T-coloring = 0 (mod 2), as each edge occurs in two of the subsets. Call this result (1).
For every S-subset of N, there are 2^(n(S)-1) different T-colorings that associate with it. That is because in each cycle of the S-subset, the two T-classes that form it will alternate, and so each cycle can be colored in two ways. Note it is 2^(n(S)-1) not 2^(n(S)) because the T-coloring is unordered.
Now, the sum of 2^(n(S)-1) * X(S) over every S of N is equivalent to summing (1) over all T-colorings which equals 0 (mod 2). 2^(n(S)-1) * X(S) is naturally 0 (mod 2) for all S for which n(S) >1. Thus it follows that sum{2^(n(S)-1) * X(S)} = 0 for when n(S) = 1, i.e. sum{X(H)} = 0. Q.E.D.
For the people that found this one here is the harder version, suppose u have 12 coins now, one of them is still conterfeit but u don't know if it's heavier or if it weight less than the others. U have 3 weighings on an accurate balance scale, find the counterfeit coint?
Arbitrarily label the coins A-L First weighing: Left A,B,C,D Right E,F,G,H If they balance, the counterfeit is in I,J,K,L. If the left is heavier, the counterfeit coin is one of A,B,C,D and it is heavier or
the counterfeit coin is one of E,F,G,H and it is lighter If the right is heavier, the counterfeit coin is one of A,B,C,D and it is lighter or
the counterfeit coint is one of E,F,G,H and it is heavier Second weighing: Case 1: Left I,J,K Right A,B,C (if known good) If they balance, coin L is counterfeit. If the left is heavier, counterfeit coin is one of I,J,K and it is heavier If the right is heavier, counterfeit coin is one of I,J,K and it is lighter Case 2: Left A,B,C,E Right D,I,J,K If they balance, counterfeit coin is one of F,G,H and it is lighter If the left is heavier, counterfeit coin is one of A,B,C and it is heavier If the right is heavier, counterfeit coin is D and it is heavier or counterfeit coin
is E and it is lighter Third weighing: If counterfeit coin is known, but not whether it is heavy or light, compare the coin
with any of the others. If counterfeit coin is X and heavy or Y and light, compare X with a good coin. If X
is heavier then X is the counterfeit, else it is Y. If counterfeit coin is heavy and one of 3 coins (X,Y,Z) Compare X with Y. If X is heavier, then X is the coin. If Y is heavier, then Y is the
4 people cross the bridge, Number one crosses in 1 min, Number two crosses in 2 min, Number 3 croses in 5 mins, Number 4 crosses in 10 mins. Now it's really dark and their scared of the dark, they have only one flashlight so they decide to go 2 by 2 to cross the bridge then one persons comes back and gives the flashlight to the others. What order must they go to cross the bridge in 17 minutes.
No. 1 and No. 2 go across: 2 minutes No. 2 returns with the flashlight: 2 minutes No. 3 and No. 4 go across: 10 minutes No. 1 returns with the flashlight: 1 minute No. 1 and No. 2 go across: 2 minutes 2 + 2 + 10 + 1 + 2 = 17 minutes
3 guys are in a hotel, they rent a room 30$ so they each pay 10 $. In the middle of the night the manager thinks 30$ is too expensive so he gives his son 5$ and tells him to go give it to the three men. The son puts 2 $ in his pocket and gives 3$ back to the three guys. So resuming this it's like if the guys paid 9X3$=27$ and their is a 2$ in the boy pocket so thats 29 in total, where did that 1$ pass from the beggining.
The equation 9x$3 = $27 is misleading. Here is an accounting of the $30 over time. Starting Time Man 1: $10 Man 2: $10 Man 3: $10 Manager: $0 Son: $0 $10+$10+$10+$0+$0 = $30 After Giving the Manager the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $30 Son: $0 $0+$0+$0+$30+$0 = $30 After Giving the Son the Money Man 1: $0 Man 2: $0 Man 3: $0 Manager: $25 Son: $5 $0+$0+$0+$25+$5 = $30 After the Son takes $2 and Gives the Men Each $1 Man 1: $1 Man 2: $1 Man 3: $1 Manager: $25 Son: $2 $1+$1+$1+$25+$2 = $30 There is no missing dollar.
suppose you have a chess board with 2 opposite corners cut. there would be 62 squares in this cut out board. you have a set of domino pieces, each piece can cover exactly 2 adjacent squares of the chess board. Is it possible to cover (tile) the cut out chess board with exactly 31 pieces of dominos? if yes, how? if not, why not?
Since two diagonally opposite squares are the same color, it leaves 30 squares of a
color and 32 of another. Since a domino only covers two squares of opposite colors,
only 15 dominos at most can be fitted on the board.
In the Protoss Lore, every time an Archon is merged, their soul is also merged [BS]. Everytime that archon dies, that souls reincarnates into a new templar following these rules: -A High Templar + High Templar archon reincarnates into a High Templar -A Dark Templar + Dark Templar reincarnates also into a High Templar -A High Templar + Dark Templar reincarnates into a Dark Templar
In the begining of Templar Time there was a known amount of each type of templar and no archons. They will merge until there is only one left. How do you determine which type of Templar will be the last remaining.
If two high templar merge, we are -1 high templar. If two dark templar merge, we are +1 high templar -2 dark templar. If one of each templar merge, we are -1 high templar. If we have an even number of dark templar to begin with, we will have an even number
of dark templar at the end. So our last templar will be a high templar. If we have an
odd number of dark templar to begin with, we will end with one dark templar at the
-Two 8-liter jugs, filled with water -One 3-liter jug, empty -Four infinite size, empty pools
Here's what your objective is: Fill each of the four pools with exactly 4 liters of water.
Let's label the jugs A, B, and C such that the first 8-liter jug is Jug A, the second
8-liter jug is Jug B, and the 3-liter jug is Jug C. Let's also label the pools pool
1, 2, 3, and 4. I was able to get a solution in 24 steps. A->C C->1 A->C A->2 C->A B->C C->A B->C C->A C->3 B->C A->C C->B A->C C->B A->C A->4 C->B C->1 B->C C->3 B->C C->4 B->2 And now all the jugs are empty and each pool has 4 liters of water.
Typical "stars" are drawn in connected, but not repeated, line segments. For example, a 5-point star is drawn as such - line segments AC, CE, EB, BD, DA. The segments must always alternate a constant number of points (in the above case, skipping 1 point in between). Given the information that there is only 1 way to draw a 5-point star, and that there is NO way to draw a 6-point star (in continuous lines, that is), and there are 2 ways to draw a 7-point star, how many different ways are there to draw a 1000-point star?
The vertices of an n-pointed star are the vertices of a regular n-gon, numbered 0
through n-1 in clockwise order. The star is determined by choosing a vertex m and
drawing the line segments from 0 to m, from m to 2m, from 2m to 3m, and (n-1)m to 0,
where all numbers are reduced modulo m. In order for the figure to satisfy our
conditions, m must be relatively prime to n and not equal to 1 or m-1. There are 400
positive numbers below 1000 that are relatively prime to 1000. Since the same star
results from choosing the first edge to go from 0 to k as when it goes from 0 to n-k,
In Madadia, a rather strange and misguided assassin, from his hidden position, uses a high-powered rifle to shoot someone in the foot from 50 feet away. The bullet travels at 1300 feet per second. Both the person being shot at and the assassin are at sea level. What will be the first evidence to the person of the attack? (As in how will he know he as been shot.)
Since the bullet travels faster than the speed of sound (1116.44 fps at sea level) he
will feel the pain of a foot thoroughly ruined before he hears the shot. So the order should be that he will see the flash, feel/realize his foot is pwnt, and then hear the shot.
Put 1001 unit squares on a coordinate plane. The squares can overlap in any fashion. Let S be the region of the plane that is covered by an odd number of squares. Prove that the area of S is greater than or equal to 1. Note: the sides of the squares are parallel to X and Y axes.
If all the squares are stacked up, then the area is one. If there is one even stack
and one odd stack, then the area is one. If there are any more than one odd stack,
then the area is greater than one. It is impossible to have no odd stacks. But a stack will have a minimum area of one. If it is an overlap, you can maximize the overlap with an odd number of squares to be all but 1 square unit, but you cannot maximize it beyond that.
To start off, a truel is exactly like a duel just with three people. One morning Mr. Black, Mr. Gray, and Mr. White decide to resolve a dispute by trueling with pistols until only one of them survives. Mr. Black is the worst shot, hitting once every three times (1/3). Mr. Gray is the second best shot, hitting his target twice out of every three times (2/3). Lastly, Mr. White always hits his target (1/1). To make it fair, Mr. Black will shot first, following by Mr. Gray (if he is still alive) and then Mr. White (provided that he is still alive). The Question is: Where should Mr. Black aim his first shot?
If Mr. Black shoots the ground, it is Mr. Gray's turn. Mr. Gray would rather shoot at
Mr. White than Mr. Black, because he is better. If Mr. Gray kills Mr. White, it is
just Mr. Black and Mr. Gray left, giving Mr. Black a fair chance of winning. If Mr.
Gray does not kill Mr. White, it is Mr. White's turn. He would rather shoot at Mr.
Gray and will definitely kill him. Even though it is now Mr. Black against Mr. White,
Mr. Black has a better chance of winning than before.
There are 3 coins on the table Gold, Silver and copper. The man at the table will let you make one statement, if it is true he will give you a coin. If it is false you won't let you have a coin. What will you say to him to always ensure that you have the gold coin?
"You will give me neither the copper or the silver coin." If it's true, you get the gold coin. If it's false, it breaks the conditions that you get no coin when lying.
On April 11 2011 12:05 x-Catalyst wrote: 4 captives are buried in the ground. There are 3 people on one side of a solid brick wall, and one person on the other. They are all facing the brick wall. The people on the left are in a line, so they can see the people in front of them. Person 1 can see person 2 and 3. Person 2 can see person 3. But person 3 and 4 can only see the brick wall in front of them. The people who are holding the captives place a star on each of the hats on their heads. 2 red, and 2 blue. They tell all 4 captives this, but do not tell them what order the stars are in. They say that if one of the captives can guess his color star, they will let him free. But if he's wrong, they all die. The captives can not talk to each other. They can only see what's ahead of them, including the other peoples stars, but not their own. The brick wall is completely solid and tall/wide. There is nothing reflective, and they cannot turn their heads to look at the people behind them. Which one of these people know what star they have, and how do they know?
The prisoners know that there are only two hats of each color. If 1 observes that 2 and 3 have hats of the same color, 1 would deduce his hat is the other color. If 2 and 3 have hats of different colors, then 1 can't say anything. 2, knowing what 1 would do, can determine that if 1 says nothing about the hats of 2 and 3, that 2's and 3's hats must be different. Being able to see 3's hat, he can determine his own hat color.
So if 2 & 3 have the same color hats, 1 will know. If 2 & 3 have different color hats, 2 will know.
You have a jar filled with 100 black marbles and a jar filled with 100 white marbles. You can rearrange the marbles any way you like as long as all of the marbles are in jars. After sorting, you randomly select one jar and then randomly select one marble from the jar. If the marble is white you win. What are your maximum odds for winning?
Place 1 white marble in one of the jar, such that it is the only marble in the jar. Place the other 99 white marbles and 100 black marbles into the other jar.
This way, you have a near 75% chance to choose a white marble.
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
Make two piles of 50 coins and flip over one of the piles. (Sounds crazy, but works!)
You have two cups with 10 teaspoons of tea in one and 10 teaspoons of coffee in the other. You take one teaspoon from the tea cup and put it in the coffee cup, then take a teaspoon of the coffee-tea mix and put it in the tea cup. Is there more coffee in the tea cup, or tea in the coffee cup?
They're the same.
Step 1 Cup 1: 10 Tea, 0 Coffee Cup 2: 0 Tea, 10 Coffee Step 2 Cup 1: 9 Tea, 0 Coffee Cup 2: 1 Tea, 10 Coffee Step 3 Cup 1: 100/11 Tea, 10/11 Coffee Cup 2: 10/11 Tea, 100/11 Coffee
Suppose you are being screened for a disease that affects 1 in 10,000 people statistically. You are to be given a test for the disease that is 99% sensitive and 99% specific, that is, that the test will correctly identify someone who has the disease as testing positive 99% of the time, and will correctly identify someone who does not have the disease as testing negative 99% of the time. How much more likely are you to have the disease given a positive test result than to not have the disease given a positive test result.
Let P(S) = the probability that you are sick (This is 0.0001 given that the incidence of disease is 1 in 10,000 people) P(N) = the probability that you are not sick (This is 1-P(N) or 0.9999) P(+|S) = the probability that the test is positive, given that you are sick (This is 0.99 since the test is 99% accurate) P(+|N) = the probability that the test is positive, given that you are not sick (This is 0.01 since the test will produce a false positive for 1% of users) P(+) = the probability of a positive test event, regardless of other information. (0.010098, which is found by adding the probability that a true positive result will appear (0.99*0.0001) plus the probability that a false positive will appear (0.01*0.9999).)
P(N|+) = the probability that a non sick person tested positive P(N|+) = P(+|N)P(N)/P(+) = 0.01*0.9999/0.010098 ~ 0.9901960784313725 P(N|+) ~ 99%
P(S|+) = the probability that a sick person tested positive P(S|+) = P(+|S)P(S)/P(+) = 0.99*0.0001/0.010098 ~ 0.0098039215686275 P(S|+) ~ 1%
P(N|+)/P(S|+) = 101.
You are 101 times more likely to NOT be sick given a positive test result than to BE sick given a positive test result. (Even though the test is 99% sensitive and 99% specific!)
You have to walk into 1 of 2 doors, where one will lead to certain instantenous death, and the other will lead to a life full of starcraft 2 and happy things. Each door is guarded by an all knowing all powerfull zergling(A crackling, if you will). One of the zerglings will always lie and the other will always tell the truth. You can ask One of them, 1 question. What do you ask to figure out what door to take? (You obviously do not know what zergling lies, and what door the zergling guards holds no relation to him lying or telling the truth.)
Ask one of the omnipotent and omniscient zerglings "If I asked the other zergling, which door would he indicate will lead to a life full of Starcraft 2 and happy things?" Then take the opposite door.
And now to pretend that no new logic problems will pop up so I don't waste my entire life here.
Let A be a collection of 100 distinct integers. Can you select 15 integers from A so that the difference of any two numbers from this selected subset is divisible by 7?
The answer is yes if the collection of 100 distinct integers is consecutive. As you
could start with the largest number in the set, and set up the recursion A(n+1) = n -
7, such that you fill a set of 15 numbers that have a difference of seven between
terms in the set. If the collection of 100 distinct integers is for example multiples
of 3s (3,6,9,12,15,18,21,etc.) then no pair of numbers subtracted will be divisible
Look at your counter example of a set. The least residue of each of those numbers mod 7 is 3, 6, 2, 5, 1, 4, and 0. These are certainly great numbers to start off with if you're trying to come up with a counter example, but unfortunately, you've just exhausted all of the least residues. The next number you pick MUST be the same least residue as one that you've already picked, and the difference between these two must be a multiple of 7 (as their difference is 0 mod 7). In your example set, the next number happens to be 24. 24 - 3 = 21, which happens to be a multiple of 7. So basically, our goal is to distribute 100 number across the 7 least residues mod 7, without any single least residue having 15 or more elements. You will find this to be impossible. We will distribute it so that each subset has 14, but that's only 98 integers, and we still have two left.
Therefore, it is ALWAYS possible to select a subset of 15 integers. You don't even need 100; you can guarantee it with 99, and they don't even have to be distinct!
Let A be a finite set of (distinct) integers. Let A+A be the set of all sums a+b where a, b are in A, and similarly define A-A to be the set of all differences a-b where a, b are in A. a and b are not necessarily distinct. Prove or disprove: |A+A| <= |A-A|.
On April 10 2011 00:05 stepover12 wrote: 40. (a round robin starcraft tournament) + Show Spoiler +
Suppose that 20 starcraft pros played in a tournament where each player competed against every other player exactly once. The result of each game is only win/lose, no draw. Is it always possible (regardless of results) to name the players 1,2,3,...,20 so that player 1 defeated player 2, player 2 defeated player 3,... player 19 defeated player 20? Give a proof or a counterexample.
Graph theory, find a Hamilton path in a directed complete graph Just look at it like you have a graph, with 20 vertices. Everyone is connection to everyone, and the connection has a direction. Basically each edges points FROM one vertex TO another one. Can be shown that for ever such complete graph there is at least one hamilton path (and the proof is also the instruction how to get it) Anyone can see that if there were only 2 players that the only edge you have actually is the path you are searching for. Either P1,P2 or P2,P1 Now the induction itself: Assume with have such a graph with n players, and we know the hamiltion path of it: P1,P2,....,Pn Now we go to n+1: We introduce a new player 0, which has either beaten or was being beaten by each of the players 1 to n Now find the player with the lowest number who was beaten by player 0, lets call this player m if he exists. If player m doesn't exist, it means EVERYONE has beaten player 0, which means you can safely put him at the end, as also player n has beaten him: P1,P2,...,Pn,P0 If player m exsists, you can insert player 0 right in front of him P1,...,P(m-1),P0,Pm,...,Pn As player m is the lowest number he has beaten, this means that player (m-1) must have beaton player 0
Four men have to cross a bridge at night. They are all on the same side. The only have one torch and only up to two men can cross the bridge at the same time. One has to have the torch to cross the bridge. Each man needs a certain time to cross the bridge. man 1: 1 min man 2: 2 min man 3: 5 min man 4: 10 min If two men cross the bridge together they take the time of the slower one. how can they cross the bridge in 17 min??
On April 11 2011 20:41 ]343[ wrote: Hmm, don't think this has been posted yet?
Let A be a finite set of (distinct) integers. Let A+A be the set of all sums a+b where a, b are in A, and similarly define A-A to be the set of all differences a-b where a, b are in A. a and b are not necessarily distinct. Prove or disprove: |A+A| <= |A-A|.
You are blind folded and presented with 100 coins. You are told that 50 are heads and 50 are tails. How can you split them into two piles so that the two piles contain an equal amount of heads? You can not distinguish the orientation of the coins by touch.
On April 12 2011 05:10 jeanphil wrote: here is one.
Four men have to cross a bridge at night. They are all on the same side. The only have one torch and only up to two men can cross the bridge at the same time. One has to have the torch to cross the bridge. Each man needs a certain time to cross the bridge. man 1: 1 min man 2: 2 min man 3: 5 min man 4: 10 min If two men cross the bridge together they take the time of the slower one.
man 1 and 2 go over : 2 mins total Man 1 goes back: 3 mins total Man 5 and 10 goes over: 13 mins total Man 2 goes back: 15 mins total Man 1 and 2 goes over: 17 mins.
On April 10 2011 00:05 stepover12 wrote: 40. (a round robin starcraft tournament) Suppose that 20 starcraft pros played in a tournament where each player competed against every other player exactly once. The result of each game is only win/lose, no draw. Is it always possible (regardless of results) to name the players 1,2,3,...,20 so that player 1 defeated player 2, player 2 defeated player 3,... player 19 defeated player 20? Give a proof or a counterexample.
Graph theory, find a Hamilton path in a directed complete graph Just look at it like you have a graph, with 20 vertices. Everyone is connection to everyone, and the connection has a direction. Basically each edges points FROM one vertex TO another one. Can be shown that for ever such complete graph there is at least one hamilton path (and the proof is also the instruction how to get it) Anyone can see that if there were only 2 players that the only edge you have actually is the path you are searching for. Either P1,P2 or P2,P1 Now the induction itself: Assume with have such a graph with n players, and we know the hamiltion path of it: P1,P2,....,Pn Now we go to n+1: We introduce a new player 0, which has either beaten or was being beaten by each of the players 1 to n Now find the player with the lowest number who was beaten by player 0, lets call this player m if he exists. If player m doesn't exist, it means EVERYONE has beaten player 0, which means you can safely put him at the end, as also player n has beaten him: P1,P2,...,Pn,P0 If player m exsists, you can insert player 0 right in front of him P1,...,P(m-1),P0,Pm,...,Pn As player m is the lowest number he has beaten, this means that player (m-1) must have beaton player 0
On April 12 2011 05:10 jeanphil wrote: here is one.
Four men have to cross a bridge at night. They are all on the same side. The only have one torch and only up to two men can cross the bridge at the same time. One has to have the torch to cross the bridge. Each man needs a certain time to cross the bridge. man 1: 1 min man 2: 2 min man 3: 5 min man 4: 10 min If two men cross the bridge together they take the time of the slower one. how can they cross the bridge in 17 min??
This one is the same as number 24 on the original post
On April 11 2011 20:41 ]343[ wrote: Hmm, don't think this has been posted yet?
Let A be a finite set of (distinct) integers. Let A+A be the set of all sums a+b where a, b are in A, and similarly define A-A to be the set of all differences a-b where a, b are in A. a and b are not necessarily distinct. Prove or disprove: |A+A| <= |A-A|.
2 DTs can merge with each-other can create an HT, lowering the DT amount by 2. One can merge with an HT, creating a DT, not changing the amount of DTs at all. Therefor, any combination of 2 DTs will not change the amount of DTs by an odd number.
Hence, if there are an odd number of DTs, one DT will remain. If there is an even number of DTs, one HT will remain.
Each star can be created by movements under half the amount of it's points (otherwise, it is the mirrored to be the exact same way as the other way around). You cannot create a star with any number that is a multiple of one of it's prime factors. You cannot create a star with movements of 1, as that will not be a star.
Therefor, 1000 cannot be made in jumps of even numbers (will skip all odd points), and numbers divisible by 5 (will skip all points not divisible by 5). So, what we are looking for is all numbers between 3 and 499 not divisible by 2 or 5. This can be discerned via the single digits of the number. In each 10, the numbers ending with 1, 3, 7 or 9 are not divisible by those numbers. there are 50 such 10s, therefor, there are 200 such numbers, needing to subtract 1 as it is not a star.
Just think of it this way: Odd overlapping causes the color to be black. Otherwise, it is empty. Adding 2 more can not subtract from this, as to subtract from it, they do an inverse on an area of between 0 (they completely overlap) and 2 (they don't overlap at all) - but no matter where there is a black area they cover, will their edges will always do an inverse to exactly the amount the overlapped with, therefor, they cannot subtract one no matter what. You start out with one square (an odd overlap with itself), and add pairs, until you reach 1001.
He should shoot not aiming at either one of the others. That way, he has a 2/3 chance of going against someone with a 2/3 chance to kill him, and otherwise, a 1/3 chance. If he kills someone else, he will most likely die the next shot.
"if you will give me the gold coin, this statement is true" - If the listener gives the gold coin, than it is true. If he does anything else, then he makes that statement true (in the "if it's snowing dogs, I can fly" kind of way), and therefor, he gives the gold coin.
If the one behind the other 2 sees 2 stars of the same color, her knows his color and leaves, and then the one in front of him knows that he is of the same color as the on in front of him and leaves as well. Otherwise, the guy in front of him knows he doesn't leave, and therefor knows he has a different color than the guy in front of him and leaves.
Therefor, only the second out of the 3 can leave each time. The other two don't have enough information, so sadly they stay in prison.
Let A be the event of picking the first jar, and B be the event in which he picked a white marble. The chances of B are the amount of white marbles out of the 100. Therefor, the chances of B happening are the chances of B if A happened, plus the chances of B if A didn't happen. Assume the first jar has M marbles. There is a 50% chance to select each jar (random choice), therefor the chance of B happening if A are M/100. Similarly, the chance of B if not A are (100-M)/100. Therefor, the chances are (1/2)*(M/100) + (1/2)*(M/100) = 1/2. Therefor, there will always be a 50% chance of picking a white marble.
Easy enough. Split them 50-50, and flip over one of the halfs. One has M heads, the other has 50-M heads before flipping over, after flipping over it will also have M heads.
First stage: cup A: 9 teaspoons of tea cup B: 10 teaspoons of coffee, 1 teaspoon of tea Second stage: Cup A: 9 tea + 10/11 coffee + 1/11 tea Cup B: 9 + 1/11 coffee + 10/11 tea
The number of diseased people is .001%, therefor, the number of healthy people is 99.99%. There is a 1% chance of a healthy person being misdiagnosed, therefor, 0.9999% of people are false positives. out of the .001%, 99% are properly diagnosed, therefor, .0099 of people are true positives. Therefor, you are 101 times more likely to get a false positive, or you have a 1/101 chance to actually have the disease given a positive result.
Does the one telling the truth guard the door that will kill me? If it is the liar, and he is protecting the bad door, he will say "yes". If it is the truth teller, and he is protecting the bad door, he will say "yes". In both cases, I go to the other door. If it is the liar and he's protecting the good door, he will say "no", if it is the truth teller protecting the good door, he will say "no", and I enter the door behind the zergling I asked the question.
Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
On April 13 2011 23:19 petitdragon wrote: |A+A| <= |A-A| + Show Spoiler +
For any 2 strictly positives and distincts integers a and b , a+b = b+a , while a-b != b-a So from there, I think we can conclude that |A+A| <= |A-A|
ah, but what if your b-a, a-b are already in the set, but a+b is not? also adding an integer a to the set adds to A+A 2a as well as a+b, for all b in A, while it adds only a-b, b-a (since 0 is already in A-A hopefully) to A-A.
It was from the Princeton Math Competition a few years back (probably 07-08?)
Let it be true for an n-1 element set (trivial for n=2).
Now given an n element set, order the elements so a_1 is the smallest and a_n is the greatest.
Let A' = A\{a_n}. We already know from induction that |A'+A'| <= |A'-A'|.
When we introduce a_n, the set of differences increases by this amount: 2(n - |B|), where B is set of pairs (n,i) such that a_n-a_i = a_j-a_k.
Notice that the set of sums increases by: n+1-|G|, where G is the set of pairs (n,i) such that a_n+a_i = a_j+a_k (j=k possible).
Now given (n,i) in G, consider a_n+a_i = a_j+a_k with a_j maximal among the possible sums. Then given (n,i) in G we can map this to the set { (n,j),(n,k) } in B. This mapping must be injective, for if we had (n,t) mapping to the set { (n,j),(n,k) } as well, then a_n + a_t = a_j+a_k would hold true, implying that i=t.
Hence we've shown that |G| => |B|. Finally note that |G| <= n-1 because the way we ordered, (n,n-1) cannot be in G. Now 2(n-|B|) > = 2(n-|G|) >= n+1 - G where the last equality is equivalent to n-1 >= G.
For |A+A| <= |A-A|; I conjecture |A+A| = (|A-A|-1)/2 + |A|, thought haven't been able to prove it. If we assume A is contained in [0,n] with 0, n in A (which is allowed as |A+A| and |A-A| are preserved under shifts of A) and call f(A) the set of positive elements of A-A, this is |A+A| = |f(A)| + |A| and it isn't too hard to show |A| <= |f(A)| + 1. As |A-A| = 2|f(A)| + 1, this would give |A+A| = |f(A)| + |A| <= 2|f(A)| + 1 = |A-A|.
For those who know a bit more math, I have a variation on #10. Suppose countably many prisoners (meaning we can assign each prisoner a natural number, ie there is a prisoner 1, prisoner 2, etc for all positive whole numbers) are given the following scenario: on the following day, an evil warden will assemble them in one (very large) room and give each of them a hat with a (not necessarily unique) natural number on it. Each prisoner will be able to see all hats except their own. Without any communication with other prisoners after receiving his or her hat, each prisoner will communicate with the warden what he or she thinks his or her number is. This communication occurs simultaneously; that is to say prisoner x has know knowledge of what prisoner y communicated to the warden (unless of course x=y...). The prisoners are all allowed free if and only if finitely many of them are wrong. Assume the prisoners know this will happen the following day, and are given a night to prepare a strategy. Is there a strategy which guarantees the prisoners all go free? Give the strategy or prove no such strategy exists.
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm..
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
if da=yes trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B) JAJA -> C is random god DAJA -> A is random god JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case) "if i were to ask you, are you the lying god, would you say da?
if da=yes trueGod: ja falseGod: da
if da=no trueGod: ja falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm..
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
if da=yes trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B) JAJA -> C is random god DAJA -> A is random god JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case) "if i were to ask you, are you the lying god, would you say da?
if da=yes trueGod: ja falseGod: da
if da=no trueGod: ja falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
Onto a puzzle of my own! You are an engineer at a light bulb company in charge of quality control and are given two identical light bulbs and a building N stories tall to test them with. You are to determine the exact floor upon which the bulbs will break (call it K, if you drop the bulbs from floor K or any floor greater than K, they break, but if you drop them from a floor less than K, they will remain intact). A broken bulb cannot be retested but an intact bulb can be. Since you are an engineer, you want to accomplish this task as fast as possible. What is the method that will guarantee a minimum maximum amount of tests needed to determine K? (Say the building has 3 floors and you drop the bulb first on floor 3, then 1, then 2. the number of tries is 3 because if K is equal to 3 or 2, it will take 3 tests to determine K. The fact that it would take two tests to determine K if K = 1 is irrelevant)
I hope this is stated clear enough so that no one needs to ask questions
I've discovered the thread yesterday and i've spent most of today working on the xkcd one and i have the solution. It's not easy, but damn it's a brilliant problem. And the solution is so brilliantly symmetric and triangular. :D
If on the island there was only one blue eyed person, then on the first night you would leave because you would see nobody else with blue eyes.
If on the island there were 2 ppl with blue eyes they would see each other, so you would not leave on the first night, because the other one could be the only blue eyed person on the island. But then you would leave on the second night, because the only possible reason the other blue eyed person didn't leave would be if he saw a blue eyed person other than himself, but you only see him with blue eyes, so you must be the other blue eyed person.
Now it gets tricky. If on the island there were 3 ppl with blue eyes then you would see 2 blue eyed people and you would imagine that each of them saw only 1. By the same reasoning as above they would leave on the second day, but they do not. Why? That's because they must see someone else with blue eyes, and that lucky guy must be you.
And so on if there were 4 ppl. You would see only 3 and they would leave on the third day, because they would follow the reasoning above. But they do not leave on the third day, there must be an extra blue eyed person, and that would be you.
Now it gets very tricky (but also unnecessary). Imagine you are one of the 100 blue eyed people. You would see 99 people who are blue eyed, and you would think that each of those blue eyed people thinks that there are 98 blue eyed people. Now imagine you are one of the 99 imaginary blue eyed people. You would see 98 blue eyed people, and you would think that each of those people saw 97 blue eyed people. Now again imagine you are one of those imaginary 98 blue eyed guys, you would see 97 blue eyed people and you would think that each of those ppl saw 96 blue eyed people. And so on.
Those imaginary people by the logic i showed before would leave after as many days as there are blue eyed people on the island. So you wait untill the 99th day and you would think that the 99 blue eyed people you see would all leave because those people imagined that there were 98 but those 98 didn't leave on the 98th day. And then you are ecstatic to find out that they don't leave! That must mean that there are 100 blue eyed people on the island, and you are one of them.
Now imagine you are a brown eyed person. And now it gets sad. You are at the 100th day are you are very nervous, because you see 100 blue eyed people and you know that if those suckers don't leave tonight, then, oh baby baby, you have a ferry ticket off the god awful silent island. Miami beach here i come! Right? And you know what? Those lucky bastards on that night pack their bags and wave you good bye. No wait, they do fucking not! They don't even communicate to you.
And to think that the only thing you need to leave the island is that someone speaks up and says: "I can see someone who has brown eyes." And yet, you see so many of them.
(Monty Hall?) On April 10 2011 02:00 Tunks wrote: How about an all time classic, just for those who haven't come across it before. Very simple if you know anything about maths though. You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice and the presenter opens one of the other 2 doors to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch?
Assume an extreme alternative: 1000 pigeons on a fence, 999 males and one female. I gotta try and find the female pigeon. I pick one with a chance of 1/1000, now the host says "let me remove some males" and he removes 998 male pigeons from the fence. Now its clear that I will have a better chance switching my choice (999/1000)
Now imagine the same experiment, but right after I picked my first pigeon, lightning strikes on the fence, and 998 pigeons drops dead from the electrocution, only 2 remains, my first pick and some other pigeon. Now the host and his helpers discovers that all the dead pigeons are in fact males. If I assume the lightning was completely independent from my first pigeon-pick, and all the dead pigeons are males, my original pigeon has now changed probability to ½, and I will not gain anything from a switch (but wont lose anything either)
I know the difference between these two scenarios has to do with conditional probability, but I´m not a mathematician so I cant explain exactly how my pigeon went from 1/1000 to ½ probability. Interesting isn't it?
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm..
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
if da=yes trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B) JAJA -> C is random god DAJA -> A is random god JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case) "if i were to ask you, are you the lying god, would you say da?
if da=yes trueGod: ja falseGod: da
if da=no trueGod: ja falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
Things get interesting when the integers are non-consecutive. This is equivalent to removing rows/columns from the matrices. Since the column and rows removed from the A+A matrix are the same that will always leave you with a symmetric matrix. The A-A will possibly be symmetric and in those cases the number of elements will be the same. But in all cases where A-A is not symmetric it will have more distinct elements than A+A.
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm..
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
if da=yes trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B) JAJA -> C is random god DAJA -> A is random god JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case) "if i were to ask you, are you the lying god, would you say da?
if da=yes trueGod: ja falseGod: da
if da=no trueGod: ja falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
Say he decides to tell the truth. He doesn't know whether he would say da or ja since it's completely up to random chance. Just because he has randomly decided to tell the truth when being asked "if i were to ask you, are you the lying god, would you say da?" doesn't mean he would tell the truth when being asked "are you the lying god." It seems to me like this second randomization breaks the problem making him unable to give an adequate answer
On April 13 2011 14:18 MusicalPulse wrote: Friend asked me this one, kinda like the other 3 person truth/lie one. Three gods A, B, and C are called, in no particular order, True, False, and Random. True always speaks truly, False always speaks falsely, but whether Random speaks truly or falsely is a completely random matter. Your task is to determine the identities of A, B, and C by asking three yes-no questions; each question must be put to exactly one god. The gods understand English, but will answer all questions in their own language, in which the words for yes and no are 'da' and 'ja', in some order. You do not know which word means which.
edit: ah, missed the fact that you don't know which is yes and which is no. maybe u do need 3 questions after all blah tricky. so far i can only figure out which is da/ja, and who is the random god in 3 questions. Hmm..
ask A "if i were to ask you, are you the random god, would you say da?". then ask B same thing. 2 questions used
if da=no trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
if da=yes trueGod: ja falseGod: ja randomGodDecidingToLie:da randomGodDecidingToTellTruth: da
therefore, in the first 2 questions we can either hear (remember we asked A then B) JAJA -> C is random god DAJA -> A is random god JADA -> B is random god
now ask one of the true/false gods (whoever they are based on case) "if i were to ask you, are you the lying god, would you say da?
if da=yes trueGod: ja falseGod: da
if da=no trueGod: ja falseGod: da
now we knew who random god was, and now we know who is who between false/true (as truegod will say Ja while falseGod will say Da)
If you ask the random god what he would say if you asked him if he was the random god or not, and he decides to tell the truth, he cannot give a "da" or "ja" answer, since because he is random, he randomly would switch between saying yes and no.
Say he decides to tell the truth. He doesn't know whether he would say da or ja since it's completely up to random chance. Just because he has randomly decided to tell the truth when being asked "if i were to ask you, are you the lying god, would you say da?" doesn't mean he would tell the truth when being asked "are you the lying god." It seems to me like this second randomization breaks the problem making him unable to give an adequate answer
huh? i don't think you get it it doesn't matter if he's lying or not, he'll end up giving the same answer due to the way the question is structured
i guess the main thing is that you're thinking he can choose to both lie AND tell the truth in order to answer this one question, but the original question, as given, explicitly forbids that. it's random as to which one he's doing, but he is doing one of them.
It's about Problem #4 (when do the clock hands (hour and minute hand) meet again after 12pm?)
Edit: Oh, wait, I just totally failed and somewhat mistook minute and second hand, zomfg x___x No surprise I mistook so many right answers for wrong ones with these incredible flaw in my idea.
Solution as I intended, with right scales: -> They won't meet again before 13:00 (1:00) for obvious reasons -> will meet before 14:00 (2:00)
Defining "1" as one time around the clock: For hour hand: 1/12 (starting point) + 1/720m (change per minute) For minute hand: 0 (starting point) + 1/60m (change per minute)
12+1/720x = 1/60m -> m = 60/11 = 5+5/11 (approximately 5 and a half) -> Hands will meet again at approx. 13:05:30
Here is the classic solution to the hour minute hand problem (I hope this is the one you were referring to):
After one revolution, the minute hand has hit the hour hand 0 times. After two revolutions, it has hit the hour hand 1 time.
So after 12 revolutions of the minute hand, it has hit the hour hand exactly 11 times. Because of even spacing, we can deduce that it takes 12/11th of an hour for the minute hand to hit the hour hand. So the next time we have an intersection is about 1 o clock + 1/11th of an hour (almost 6 minutes)
Of course, this is just a prettier restatement of the algebraic solution: In one hour, the hour hand moves 30 degrees while the minute hand moves 360 degrees. Let x be the number of hours so we should write: 360+30x = 360x or x = 360/330 = 12/11 of an hour.
Hello gents! I'm having some trouble solving a problem and I hope someone of you fine people can shed some light over it. It probably require that you have studied some mathematics. Determine a positive integer and a negative integer such as both integers by division with 153 gives the rest/remainder 7 and such as both integers by division with 177 gives the rest/remainder 4.
I'm sorry if it doesn't make any sense but I've translated the text from 4 different language and I finally understand the meaning of, lost in translation.
On April 15 2011 14:14 noob4ever wrote: but I´m not a mathematician so I cant explain exactly how my pigeon went from 1/1000 to ½ probability. Interesting isn't it?
It didn't.
That situation is identical to selective removal of pigeons.
OK one of my favorite problems (hopefully not posted here already--it's somewhat of a classic).
Find all partitions of the positive integers into 2 sets S and T such for any positive integer n, there are exactly as many ways of expressing it as the sum of 2 distinct integers from S as there are from T.
Ok, ever since I've found this thread I've been trying to remember some puzzles, here are a few...
An odd number of soldiers are stationed in a field, in such a way that all the pairwise distances between all soldiers are distinct. Each soldier is instructed to watch the soldier closest to himself. Prove that at least one soldier is not being watched.
(Proof by contradiction): Consider the two soldiers at the shortest distance from each other. They are watching each other; if anyone else is watching one of them, then we have a soldier being watched twice (and therefore another would not be watched). Otherwise, we could remove the pair from the field without affecting the watching patterns of the others. Continuing in this manner, we would be left with one soldier left in the field, not watching anyone, the contradiction.
Mike and Jane go to a dinner party with four other couples; each person there shakes hands with everyone he or she doesn't know. Later, Mike does a survey and discovers that every one of the nine other attendees shook hands with a different number of people. How many people did Jane shake hands with?
The key here is that Jane is the partner of the one person NOT polled. Since each person shook hands with at most 8 people, the answers Mike receives are one each from the set [0, 1,..., 8]. Consider a couple named A and B, who scored 8 and 0, respectively. They must have been partners, since otherwise the opportunity to shake each other's hands would've ruined their scores. Consider C and D, as well, with scores 1 and 7. They too must be a couple, since C shook A's hand, and D must've missed B; if C and D weren't couples, their scores would be different. Similarily, the scores 2-6 and 3-5 must be couples, leaving Mike and Jane shaking hands with only the high scorers. Therefore, both Mike and Jane shook hands with 4 people.
EPT
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but the real answer you seek is some point near 1:05 that i am too lazy to calculate. you just have to remember it isn't 1:05 because the hour hand will have moved up 1/12 of the distance between 1 and 2 at that point.
9 B:0 C:1. my reason for this is that A wants to keep as much of it as possible to himself and since he needs another vote to survive he gives the pirate with no chances of gaining money if he dies a piece of the loot in order to get the needed votes, i dont think it will ever come to only 2 remaining pirates.
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