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Gollum challenges you to a contest of riddles
In the same vein as the Horror Discussion thread, I want to make a thread for interesting logical riddles to see if you guys on TL can solve them. Of course, post your own riddles too, I'll update the OP every once in a while with good riddles that others have posted, and with others I find. Bolded ones I personally enjoyed a lot.
Edit: Added answers
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Marble Jars [tag: math, probability, logic] + Show Spoiler +You are a prisoner in a foreign land. your fate will be determined by a little game. There are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)? Answer: + Show Spoiler +1 white marble in one jar, all 99 of the other marbles in the other jar. 3 Hats [tag: logic] + Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be? Non Homogeneous Rope Burning[tag: math, logic] + Show Spoiler +You have two ropes, each of which takes one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using these non-homogeneous ropes and a lighter, time 45 minutes. Note: Some clarification on what is meant by non-homogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaining 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off. Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox] + Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. Footsize and Spelling Ability [tag: thinkoutsidethebox] + Show Spoiler +Scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests. How can you explain this correlation? Answer: + Show Spoiler + Cork, Bottle, Coin [tag: thinkoutsidethebox] + Show Spoiler +If you were to put a coin into an empty bottle and then insert a cork in the bottle's opening, how could you remove the coin without taking out the cork or breaking the bottle? Answer: + Show Spoiler +Push the cork into the bottle Glass Half Full [tag: math] + Show Spoiler +You are in an empty room and you have a transparent glass of water. The glass is a right cylinder, and it looks like it's half full, but you're not sure. How can you accurately figure out whether the glass is half full, more than half full, or less than half full? You have no ruler or writing utensils. Hint: + Show Spoiler + tagged math because geometry is math. Answer: + Show Spoiler +tilt it so that the water is just shy of spilling out the lip of the glass, then observe where the water level is in relation to the bottom of the glass Faustian Round Table Game [tag: logic] + Show Spoiler +You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a round table. He gives himself and you a huge pile of quarters. He says "ok, we'll take turns putting quarters down, no overlapping allowed, and the quarters must rest on the table surface. the first guy who can't put a quarter down loses." you guys are about to start playing, and the devil says that he'll go first. however, at this point you immediately interject, and ask if you can go first instead. you make this interjection because you are very smart, and you know that if you go first, you can guarantee victory. explain how you can guarantee victory. Note: If you can put a quarter down so that it balances on the edge, then so can the devil Answer: + Show Spoiler +placing the first coin in the exact center of the table gives one the strategy of matching the devil's every other move on the opposite side of the table Red Eyes and Blue Eyes [tag: logic] + Show Spoiler +There is an island of monks where everyone has either brown eyes or red eyes. Monks who have red eyes are cursed, and are supposed to commit suicide at midnight. However, no one ever talks about what color eyes they have, because the monks have a vow of silence. Also, there are no reflective surfaces on the whole island. Thus, no one knows their own eye color; they can only see the eye colors of other people, and not say anything about them. Life goes on, with brown-eyed monks and red-eyed monks living happily together in peace, and no one ever committing suicide. Then one day a tourist visits the island monastery, and, not knowing that he's not supposed to talk about eyes, he states the observation "At least one of you has red eyes." Having acquired this new information, something dramatic happens among the monks. What happens? Where's the father? [tag: math, thinkoutsidethebox] + Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? Answer: + Show Spoiler +The math works out to -3/4, so the baby is 9 months shy of being born, so the father is in the mother. Three lightbulbs[tag: thinkoutsidethebox] + Show Spoiler +You are in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You are allowed only one chance to enter the room with the light bulbs. How can you determine which lightswitch goes with which light bulb? Answer: + Show Spoiler +Turning on a lightbulb introduces a third variable, heat. Heat doesn't dissipate right away like light does. Manholes + Show Spoiler +Why are manholes round? Answer: + Show Spoiler +So they don't fall into the sewer like a square one could. Forcefield Detainment [tag: logic] + Show Spoiler +A group of prisoners are trapped in a forcefield. These prisoners are perfectly brave, meaning that they would attempt an escape on any positive probability of success. The prisoners are monitored by a guard who has only one bullet in his gun, but who also has perfect marksmanship skills (he never misses). A maintenance technician needs to tune up the forcefield generator, and so for one second, the forcefield is released. How can the guard still keep all the prisoners detained? Note: The prisoners are amoral, so they will back out of agreements with each other made a priori if they are sure they will be the one dying. Also, assume no prisoner can be manhandled out of the forcefield radius by the other prisoners. Answer: + Show Spoiler +The guard labels the prisoners each with a number, and issues the statement "If any of you try to escape, I will shoot the prisoner with the highest number trying to escape." With this knowledge, all will chicken out Lemming Drownings[tag: logic] + Show Spoiler +Somewhere in Northern Eurasia, a group of 20 lemmings is planning a special group suicide this year. Each of the lemmings will be placed in a random position along a thin, 100 meter long plank of wood which is floating in the sea. Each lemming is equally likely to be facing either end of the plank. At time t=0, all the lemmings walk forward at a slow speed of 1 meter per minute. If a lemming bumps into another lemming, the two both reverse directions. If a lemming falls off the plank, he drowns. What is the longest time that must elapse till all the lemmings have drowned? Hint: + Show Spoiler +After making a certain observation, you'll find the calculations trivial Answer: + Show Spoiler +100 minutes, assuming 1 or more are placed at the very end of the board. The observation one needs to make is that bumping into each other and each changing directions is no different from sliding past each other untouched Infinite Quarters [tag: math, logic] + Show Spoiler +You are wearing a blindfold and thick gloves. A very large, but finite, number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all? Answer: + Show Spoiler +Starting with all the coins in one pile, move 20 coins to another pile, flip all of them. You win. 12 balls [tag: math, logic] + Show Spoiler +You have 12 identical-looking balls. One of these balls has a different weight from all the others. You also have a two-pan balance for comparing weights. Using the balance only 3 times, how can you determine which ball has the unique weight, and also determine whether it is heavier or lighter than the others? Answer (fucking long ass answer, I just copy and pasted someone who wrote it out on a forum): + Show Spoiler +Divide the balls into three groups of four each, AAAA, BBBB and CCCC. Weigh AAAA-BBBB. Possible results are: - They balance: Means one of the C's is heavy or light. Therefore, weigh CCC-AAA (all A's are now known to be standard): They balance: Means the 4th C is the oddball. Therefore, weigh the 4th C against any other ball. 4th C falls: Means The 4th is heavy. 4th C rises: Means The 4th is light. CCC side falls: Means one of the C's is heavy. Therefore, weigh C-C: They balance: Means The other C is heavy. One side falls: Means That C is heavy. CCC side rises: Means one of the C's is light. Therefore, weigh C-C. They balance: Means the other C is light. One side rises: Means that C is light.
The AAAA side falls: Means the oddball is either a heavy A or a light B and the C's are all standard. Therefore, arrange the balls into three new groups like so: AAAC BBBA CCCB. Weigh BBBA-CCCB: They balance: Means Oddball in AAAC. Therefore, weigh A-A. They balance: Means the other A in AAAC is heavy. One side falls: Means that side is the heavy A. Left side (BBBA) falls: Means the A in BBBA is heavy or the B in CCCB is light. Therefore, weigh A-C (C is known to be standard). They balance: Means The B in CCCB is light. A side falls: Means A is heavy. C falls: Not possible. Right side (CCCB) falls: Means a B in BBBA is light. Therefore, from BBBA weigh B-B. They balance: Means the other B in BBBA is light. Left side falls: Means the B on the right is light. Right side falls: Means the B on the left is light.
The BBBB side falls: the oddball is either a heavy B or a light A and the C's are all standard. Therefore, arrange the balls into three new groups like so: AAAB BBBC CCCA. Weigh AAAB-CCCA: They balance: Means the oddball is in BBBC. Therefore, weigh B-B. They balance: Means the other B in BBBC is heavy. One side falls: Means that side is the heavy B. Left side (AAAB) falls: The B in AAAB is heavy or the A in CCCA is light. Therefore, weigh B-C (C is known to be standard). They balance: Means the A in CCCA is light. B side falls: Means B is heavy. C side falls: Not possible. Right side (CCCA) falls: Means an A in AAAB is light. Therefore, from AAAB weigh A-A. They balance: Means the other A in AAAB is light. Left side falls: Means the A on the right is light. Right side falls: Means the A on the left is light. Coin Machine Weighing [tag: math, logic] + Show Spoiler +you have 20 coin machines, each of which produce the same kind of coin. you know how much a coin is supposed to weigh. one of the machines is defective, in that every coin it produces weighs 1 ounce less than it is supposed to. you also have an electronic weighing machine. how can you determine which of the 20 machines is defective with only one weighing? (by one use, we mean you put a bunch of stuff on the machine and read a number, and that's it -- you not allowed to accumulate weight onto the machine and watch the numbers ascend, because that's just like multiple weighings). you are allowed to crank out as many coins from each machine as you like. Answer: + Show Spoiler +Take one coin from the first machine, 2 from the second, three from the third... etc. Assuming each coin weighs n ounces, read how many ounces less than 210n the scale give you. ===============================================================
Note: Most of these ones I started out with are collected by a person called William Wu, from UC Berkley. His website has tons of good riddles, but a lot of them are programming/chess/hardcore math related too. I picked out some of the ones I liked more form his site to kick off this thread.
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On December 05 2010 13:12 t3tsubo wrote:Marble Jars [tag: math, probability, logic]+ Show Spoiler +You are a prisoner in a foreign land. your fate will be determined by a little game. There are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)? Don't forget spoilers guys + Show Spoiler +I think it's a 50% chance of living no matter what you do. If you don't redistribute the jars at all, you will either be given a jar with all white marbles (100% chance of living) or with all black marbles (0% chance of living). When you average this out you get a 50% chance.
The other border case is when you make it so both jars are 50% white and 50% black. In either case, you have a 50% chance of picking a white marble and living.
Were I given this choice, and I weren't allowed to go kick-ass on the natives (feeling merciful that day), I'd just put all the black marbles at the bottom. Even though the jars are shaken up, unless they are inverted and shaken for hours, my chance of picking a white marble is still significantly larger from the top.
3 Hats [tag: logic]+ Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be?
+ Show Spoiler + C knows he is wearing a black hat.(Assuming A, B, and C are math/engineering majors)(not statisticians)
This question is quite simple and you can just brute force through every single combination of outcomes and check if they are logical.
First let's start with A, the logical choice. He can't be seeing two white hats, because if he did he would know that he were wearing a black hat. That means that he either sees a black and a white hat, or he sees two black hats.
Now we ask, well what if B saw a white hat. Well then A could not have seen two black hats and so he would know that A had seen a white and black hat (the only other possibility). C would have been wearing white and B, himself would have known that he were wearing a black hat. But he did not know which color hat he were wearing.
Therefore B saw a black hat on C. C knows this. C is wearing a black hat.
Moar plox, these are fun!
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United States1719 Posts
I love riddles so much Solutions to some of them are below. Meanwhile, here's some riddles of my own that I picked up over the years!
The apple truck and apple-eating squirrel + Show Spoiler +There is an apple truck that can hold up to 1000 apples at any given time. It needs to make a delivery of 4000 apples from town A to B, which is 900 miles apart. However, there is a squirrel living in the truck that will eat 1 apple per mile. for example, if you load 1000 apples and take 4 trips, you will end up with 4x100 = 400 apples left. Try to maximize the number of apples you can deliver. Hint: + Show Spoiler +You can leave apples on the road along the way, and pick them up whenever you want, assuming your truck has the capacity to hold them.
The blind man and checker board + Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1
The mars robots (Some computer science knowledge required) + Show Spoiler +NASA is trying to launch two robots onto mars so they can construct a station. This is a two robot job, so both robots must work together to build the station. However, they do not have a spaceship big enough to fit both, so decide to launch them separately. Although both rockets will be following the same launch protocols, some error will be inevitable and the robots will land in different positions. Now assume that mars is an endless array of trillions and zillions of squares, stretched side-by-side. Basically, it is an infinite one-dimensional space. Your goal is to program the same algorithm on both robots so when they land, they will start looking for each other, and construct the station together. The robots CANNOT be programmed with different algorithms, and have no means of communication to earth or each other or any entity whatsoever. However, the robots will deploy a parachute when landing, and that parachute will remain in the square they land forever. The robot can detect whether they have found a parachute in the square it is in. Come up with an algorithm so that they will be guaranteed to find each other. (Normal restrictions in programming apply, such as integer overflow: any number larger than 2^32 will be undefined, so if you are keeping a counter and it reaches above that, it will render your algorithm useless.)
Solutions to t3tsubo's riddles:
On December 05 2010 13:17 Hidden_MotiveS wrote:Show nested quote +On December 05 2010 13:12 t3tsubo wrote:Marble Jars [tag: math, probability, logic]+ Show Spoiler +You are a prisoner in a foreign land. your fate will be determined by a little game. There are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)? Don't forget spoilers guys + Show Spoiler +I think it's a 50% chance of living no matter what you do. If you don't redistribute the jars at all, you will either be given a jar with all white marbles (100% chance of living) or with all black marbles (0% chance of living). When you average this out you get a 50% chance.
The other border case is when you make it so both jars are 50% white and 50% black. In either case, you have a 50% chance of picking a white marble and living.
Were I given this choice, and I weren't allowed to go kick-ass on the natives (feeling merciful that day), I'd just put all the black marbles at the bottom. Even though the jars are shaken up, unless they are inverted and shaken for hours, my chance of picking a white marble is still significantly larger from the top.
+ Show Spoiler +distribute it so 1 jar has just 1 white marble, and the other jar has all other 99 marbles. Chance to live is (0.5 * 1) + (0.5 * 49/99) which I am too lazy to calculate, but certainly more than 50%
3 Hats [tag: logic]+ Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be?
+ Show Spoiler +C is wearing black, because if he were wearing white, either A or B would have known the color of their own hat. So if C were wearing white, there would be two cases: 1) where B is also wearing white, and 2) where B is wearing black. In case 1), A would have known his own hat is black, because there are only 2 white hats available. In case 2), where B is wearing black, B would be able to infer that his own hat is black from A's comment, because if it were not, A would have known his hat color. Therefore, there are no cases to draw those answers if C were wearing white.
Non Homogeneous Rope Burning[tag: math, logic]+ Show Spoiler +You have two ropes, each of which takes one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using these non-homogeneous ropes and a lighter, time 45 minutes. Note: Some clarification on what is meant by non-homogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaining 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off.
+ Show Spoiler +You would start burning both ends of one rope, and one end of the other rope, all at the same time. By the time the rope with both ends on fire burns out (30 minutes) light the other end of the other rope on fire too. The remaining rope still has 30 minutes worth of burning left, and lighting both ends will cause it to burn in half the time, which is 15 minutes. When that rope finishes burning, 30minutes+15minutes = 45minutes will have passed.
Footsize and Spelling Ability [tag: thinkoutsidethebox]+ Show Spoiler +Scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests. How can you explain this correlation? + Show Spoiler +The correlation has an external variable: age. The older the kid is, the bigger his footsize tends to be, and also tends to have gone through more education/school. Thus, footsize and performance in spelling bees/spelling tests are unrelated to each other, just proportional to age.
Glass Half Full [tag: math]+ Show Spoiler +You are in an empty room and you have a transparent glass of water. The glass is a right cylinder, and it looks like it's half full, but you're not sure. How can you accurately figure out whether the glass is half full, more than half full, or less than half full? You have no ruler or writing utensils. Hint: + Show Spoiler + tagged math because geometry is math. + Show Spoiler +you tilt the glass so that the surface of the water forms a diagonal from the rim of the open ended side to the top corner in the close-ended side. If it forms a perfect line that extends from the circumference of the bottom circle to that of the top circle's, there's exactly half left.
Faustian Round Table Game [tag: logic]+ Show Spoiler +You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a round table. He gives himself and you a huge pile of quarters. He says "ok, we'll take turns putting quarters down, no overlapping allowed, and the quarters must rest on the table surface. the first guy who can't put a quarter down loses." you guys are about to start playing, and the devil says that he'll go first. however, at this point you immediately interject, and ask if you can go first instead. you make this interjection because you are very smart, and you know that if you go first, you can guarantee victory. explain how you can guarantee victory. Note: If you can put a quarter down so that it balances on the edge, then so can the devil
+ Show Spoiler +put a coin in the center of the circle, then just make a mirror image of whatever the devil does.
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Ares[Effort]
DEMACIA6550 Posts
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Here's one of my favourites. Logic:
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United States1719 Posts
On December 05 2010 13:30 Dr. ROCKZO wrote:Here's one of my favourites. Logic: so does each candy in the poisoned bag weigh more? as in: each piece of candy in the other three bags weight 5kg/100 = 50g, whereas candy in the poisoned bag weight 5.01kg/100 = 50.1g? If that assumption is wrong please clarify for me hehe
On December 05 2010 13:30 Dr. ROCKZO wrote: Yes, I think that is what it is supposed to be but the question doesn't say that so if that's the case it is a very poor version of the riddle
Based on this assumption, here's a solution:+ Show Spoiler +arbitrarily label each bag 1, 2, 3 and 4. Take 1 candy from bag 1, 2 from bag 2, and so on, so you have 10 pieces. Measure it on the scale, and the number in the first decimal digits will tell you which bag is the poisoned bag. ex) 500.4g = bag 4 is poisoned, and so on.
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Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I?
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Non Homogeneous Rope Burning[tag: math, logic] + Show Spoiler +You have two ropes, each of which takes one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using these non-homogeneous ropes and a lighter, time 45 minutes. Note: Some clarification on what is meant by non-homogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaining 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off.
+ Show Spoiler +Start by burning both ends of Rope A and 1 end of Rope B. When Rope A is completely burned 30 minutes will have passed and there will be 30 minutes left of Rope on Rope B. Start burning the other end of Rope B and that should take 15 minutes to burn, giving you 45.
Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox] + Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours)
+ Show Spoiler +He can start a fire at the B side of the island and hope that it burns off before the fire at A side gets to him. Then he can stand in the ashes of the 2nd fire and the A fire will have no more fuel to get to him. Or he could learn to swim or build a boat before the fire gets to him.
Footsize and Spelling Ability [tag: thinkoutsidethebox] + Show Spoiler +Scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests. How can you explain this correlation?
+ Show Spoiler +Bigger foot size means they are older. A 10 year old outspells a 5 year old anyday
Cork, Bottle, Coin [tag: thinkoutsidethebox] + Show Spoiler +If you were to put a coin into an empty bottle and then insert a cork in the bottle's opening, how could you remove the coin without taking out the cork or breaking the bottle?
+ Show Spoiler +Push the cork into the bottle and you're technically not "taking out the cork"
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Faustian Round Table Game [tag: logic]+ Show Spoiler +You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a round table. He gives himself and you a huge pile of quarters. He says "ok, we'll take turns putting quarters down, no overlapping allowed, and the quarters must rest on the table surface. the first guy who can't put a quarter down loses." you guys are about to start playing, and the devil says that he'll go first. however, at this point you immediately interject, and ask if you can go first instead. you make this interjection because you are very smart, and you know that if you go first, you can guarantee victory. explain how you can guarantee victory. Note: If you can put a quarter down so that it balances on the edge, then so can the devil + Show Spoiler +If you put it in the very middle first, each subsequent turn then mirror the devil along any of the axes of symmetry. As long as he can play, there will be a symmetrical spot for you to play.
Of course you will still lose because the devil has supernatural powers and can place them perfectly whereas you are going to have a hard time measuring the exact symmetrical spot.
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On December 05 2010 13:36 rotinegg wrote:Show nested quote +On December 05 2010 13:30 Dr. ROCKZO wrote:Here's one of my favourites. Logic: so does each candy in the poisoned bag weigh more? as in: each piece of candy in the other three bags weight 5kg/100 = 50g, whereas candy in the poisoned bag weight 5.01kg/100 = 50.1g? If that assumption is wrong please clarify for me hehe
+ Show Spoiler +Yes, I think that is what it is supposed to be but the question doesn't say that so if that's the case it is a very poor version of the riddle
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On December 05 2010 13:42 BlackJack wrote:Footsize and Spelling Ability [tag: thinkoutsidethebox]+ Show Spoiler +Scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests. How can you explain this correlation? + Show Spoiler +Bigger foot size means they are older. A 10 year old outspells a 5 year old anyday + Show Spoiler +No it's obviously boys are smarter than girls
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Glass Half Full [tag: math]
+ Show Spoiler + A half filled right cylinder should have the property so that if you were to hold the cup diagonally, you can adjust it so that the liquid touches the lip and the edge of the bottom (easier to picture it in 2 dimensions: simply a rectangle cut diagonally into two equal triangles
Faustian Round Table Game [tag: logic]
+ Show Spoiler +Place a coin in the middle, mirror every move the devil does directly across
Where's the father? [tag: math, thinkoutsidethebox]
+ Show Spoiler +Math amounts to about -0.6 years, which is approximately 8-9 months, meaning the dads boning the mother
Red Eyes and Blue Eyes [tag: logic]
+ Show Spoiler + Case of 1 monk with red eyes: He will see that everyone else has brown eyes and will kill himself the first night. Case of 2 monks with red eyes: No one kills themselves the first day, and so the red eyed monks realize that there must be 2 monks with red eyes, whoever sees only one other monk with red eyes will kill himself. Case of 3 monks: No one kills themselves the first two days, monk with red eyes realize that there are 3 monks with red eyes while they only see 2 other monks with red eyes, and all monks with red eyes kill themselves. Case N: No one kills themselves for the first N-1 days, monks with red eyes realize that there are N monks with red eyes while they only see N-1 monks with red eyes, and all monks with red eyes kill themselves.
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Cork, Bottle, Coin [tag: thinkoutsidethebox] + Show Spoiler +If you were to put a coin into an empty bottle and then insert a cork in the bottle's opening, how could you remove the coin without taking out the cork or breaking the bottle?
+ Show Spoiler +use a cork that is too small for the opening of the bottle, so that the cork goes all the way inside the bottle
are you going to post answers?
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bags o candy with poison involved ^^
+ Show Spoiler +okay, so heres my solution. IF IF IF IF all the candies in each individual bag are the same weight!!! + Show Spoiler +
100 candy per bag, 5000 grams of weight per bag, 5010 grams of weight for poison bag
so a regular candy is 50g per, while a poison candy is 50.1g per
let's mark bags ABCD
take 1 piece from bag A 2 pieces from bag B 3 pieces from bag C
place them on the scale, seperated enough to remember which came from which
if the scale reads....
300g (50g x6) then bag D is poison 300.1g (50g x5, 50.1g x1) then bag A is poison 300.2g (50g x4, 50.1g x2) then bag B is poison 300.3g (50g x3, 50.1g x3) then bag C is poison
then you replace whatever candies still are edible and enjoy the candy
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Canada8025 Posts
Where's the father? [tag: math, thinkoutsidethebox]+ Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? + Show Spoiler +The math doesn't work out, with the child's age being something ridiculous like -3/4. So obviously something's funny.
So let's say that when the mother was 22, the child was 1 year old. Thus, the initial condition that the mother is 21 years older than the child is met.
Three years later the initial condition, the mother dies. She is 25, and the child is 4. Four years after the initial condition, the child dies. So six years after the initial condition, we can say that the mother is 25 and the child is 5, so she is 5 times as old as the child.
So my answer is that the father is alone.
On December 05 2010 13:38 wishbones wrote: Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I? + Show Spoiler +The man. Hence "that man's father" is your father, and the "father's son" is yourself.
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I'm still trying to figure this one out (from Ares[Effort]'s riddle thread):
On March 16 2009 14:27 l10f wrote: The MBC Game Hero team decided to go on a trip to the USA. Light decided to bring his laptop to practice with him. All the other players were jealous that he could practice while they couldn't. One day, his laptop went missing while he slept.
1. Jaehoon claims that he was busy hooking up with girls last night. 2. Shark says that he never even saw the laptop. 3. Pusan says he saw Sea take the laptop. 4. Sea says he was out last night on a one night stand with a girl he met, and that Pusan is lying 5. Saint says Shark was in Light's room at 3AM last night.
Any of the players could be telling a lie, or could be telling the truth. Who stole Light's laptop?
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On December 05 2010 14:11 Spazer wrote:Show nested quote +Where's the father? [tag: math, thinkoutsidethebox]+ Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? + Show Spoiler +The math doesn't work out, with the child's age being something ridiculous like -3/4. So obviously something's funny.
So let's say that when the mother was 22, the child was 1 year old. Thus, the initial condition that the mother is 21 years older than the child is met.
Three years later the initial condition, the mother dies. She is 25, and the child is 4. Four years after the initial condition, the child dies. So six years after the initial condition, we can say that the mother is 25 and the child is 5, so she is 5 times as old as the child.
So my answer is that the father is alone.
You finished the math part of the riddle correctly (-3/4), now you just have to do the thinkoutsidethebox part.
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On December 05 2010 14:11 Spazer wrote:Show nested quote +Where's the father? [tag: math, thinkoutsidethebox]+ Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? + Show Spoiler +The math doesn't work out, with the child's age being something ridiculous like -3/4. So obviously something's funny.
So let's say that when the mother was 22, the child was 1 year old. Thus, the initial condition that the mother is 21 years older than the child is met.
Three years later the initial condition, the mother dies. She is 25, and the child is 4. Four years after the initial condition, the child dies. So six years after the initial condition, we can say that the mother is 25 and the child is 5, so she is 5 times as old as the child.
So my answer is that the father is alone.
+ Show Spoiler +If you got -3/4, the 'baby child' has just existed, and -3/4 is also 9 months, so we can conclude the father is in the mother...
How do you get -3/4? 21+c=m, 5c=m+6?
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Marble Jars [tag: math, probability, logic] + Show Spoiler +You are a prisoner in a foreign land. your fate will be determined by a little game. There are two jars, one with 50 white marbles, and one with 50 black marbles. at this point, you are allowed to redistribute the marbles however you wish (e.g. swap a black marble with a white marble, etc.): the only requirement is that after you are done with the redistribution, every marble must be in one of the two jars. afterwards, both jars will be shaken up, and you will be blindfolded and presented with one of the jars at random. then you pick one marble out of the jar given to you. if the marble you pull out is white, you live; if black, you die. how should you redistribute the marbles to maximize the probability that you live; what is this maximum probability (roughly)? Uhh yeah I think I've found the answer to the first one, it's not 'the same either way' lol. + Show Spoiler + 1 white marble in 1 jar, 49 white and 50 black in another. This essentially gives you 1/2 +49/99 chance of survival which is almost 2/3
3 Hats has been done above and Yes, I've come to the same conclusion. It's a rather old riddle (there is a continuation of this whereby we have 10 people, A through I, who can only live if they can guess the colour of their own hats, the riddle being how many of them can be saved?)
Red Eyes and Blue Eyes [tag: logic] + Show Spoiler + There is an island of monks where everyone has either brown eyes or red eyes. Monks who have red eyes are cursed, and are supposed to commit suicide at midnight. However, no one ever talks about what color eyes they have, because the monks have a vow of silence. Also, there are no reflective surfaces on the whole island. Thus, no one knows their own eye color; they can only see the eye colors of other people, and not say anything about them. Life goes on, with brown-eyed monks and red-eyed monks living happily together in peace, and no one ever committing suicide. Then one day a tourist visits the island monastery, and, not knowing that he's not supposed to talk about eyes, he states the observation "At least one of you has red eyes." Having acquired this new information, something dramatic happens among the monks. What happens?
+ Show Spoiler + This one's pretty hard. If there is only one red-eye'd monk, then he would commit suicide on the first night. If nobody has committed suicide on the first night, that means there are two or more red-eyed monks. If there are two, then they will realise by the second night and two monks will commit suicide (because each will realise and only count one after the first day, the one that has not committed suicide yet). If this does not happen, then there are at least three, and if there are, then those will commit suicide on the third night, and so on, and so forth, until the night that there are that many red-eyed monks as days that have passed, when they will all commit suicide. Now I am sad because monks have committed suicide =(
Where's the father? [tag: math, thinkoutsidethebox] + Show Spoiler + The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father?
+ Show Spoiler +He's in bed with the mother (Do the math, the child's age is -3/4 years, which is exactly 9 months. He is still in the womb and has just been conceived. Obviously, the 'father' has just impregnated the 'mother', even though by conventional terms, we would not refer to them as thus)
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Canada8025 Posts
On December 05 2010 14:22 Blisse wrote:Show nested quote +On December 05 2010 14:11 Spazer wrote:Where's the father? [tag: math, thinkoutsidethebox]+ Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? + Show Spoiler +The math doesn't work out, with the child's age being something ridiculous like -3/4. So obviously something's funny.
So let's say that when the mother was 22, the child was 1 year old. Thus, the initial condition that the mother is 21 years older than the child is met.
Three years later the initial condition, the mother dies. She is 25, and the child is 4. Four years after the initial condition, the child dies. So six years after the initial condition, we can say that the mother is 25 and the child is 5, so she is 5 times as old as the child.
So my answer is that the father is alone. + Show Spoiler +If you got -3/4, the 'baby child' has just existed, and -3/4 is also 9 months, so we can conclude the father is in the mother... Ohhhh, that's pretty clever. Better than my morbid version.
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you should post answers =/
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added a poll on whether I should include answers in the OP
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Bosnia-Herzegovina114 Posts
Burning island of doom Solution: + Show Spoiler +He needs to start his own fire ahead of the original one and follow it at a leisurely pace of ... ugh I don't know why you put the windspeed because I don't know if I'm supposed to add windspeed to it or not, however if the secondary fire is affected by wind so is the primary, but you said it will take 10 hours for it to burn the whole island, so it can't be. Anyways, that's the answer, minutia is less important. I remember reading a riddle with the same solution as a kid in a book with conundrums and magic tricks called something like "Little smartypants" :-)
Where's the father? Solution: + Show Spoiler +On the mother, without a condom! :-) x = 21+y x+6 = 5y+30 This gives y = -3/4 (or -9 months, since we're talking years)
Glass Half Full Solution: + Show Spoiler +You tip the glass towards yourself, so the water is almost spilling out. Now observe where the water level ends: if it covers only the bottom of the glass, then it's exactly half full (or empty, as you wish). If the bottom isn't entirely covered, it's less and if it covers the walls then it's more than half. Again, something similar to this was in the same book I mention above, only this time you were going to the market to buy a barrel of wine and the guy selling it swears there is exactly half a barrel left. How will you check if he's telling the truth? Now that I've explained ... day9? Where did you come from? We're doing riddles, come on man, put the glass down ... Sheesh :-)
3 hats Solution: + Show Spoiler +Do you really need a solution? The C dude is wearing a black hat
Cork, Bottle, Coin Solution: + Show Spoiler +Ok, usually there is only one solution, but this one has infinite possibilities (I'd blame the wording). For example, riddle doesn't explicitly forbid slicing the bottle in half with a katana, now does it? :-)
Edit: removed those sneaky extra tags
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On December 05 2010 14:13 Aeres wrote:I'm still trying to figure this one out (from Ares[Effort]'s riddle thread): Show nested quote +On March 16 2009 14:27 l10f wrote: The MBC Game Hero team decided to go on a trip to the USA. Light decided to bring his laptop to practice with him. All the other players were jealous that he could practice while they couldn't. One day, his laptop went missing while he slept.
1. Jaehoon claims that he was busy hooking up with girls last night. 2. Shark says that he never even saw the laptop. 3. Pusan says he saw Sea take the laptop. 4. Sea says he was out last night on a one night stand with a girl he met, and that Pusan is lying 5. Saint says Shark was in Light's room at 3AM last night.
Any of the players could be telling a lie, or could be telling the truth. Who stole Light's laptop? That's because there's absolutely no information given here that could help you figure it out.
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added a few more riddles since the solution to all of the ones I posted first have already been found out~
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***WARNING*** My answers in spoilers
Marble Jars [tag: math, probability, logic] - 2 minutes + Show Spoiler +Put 1 white marble in one jar. Now put 49 white marbles and 50 black marbles in the other jar. Odds of living ~ .5(1) + .5(.49495) ~ 74.4747475%
3 Hats [tag: logic] - 10 minutes + Show Spoiler +Person C is black-hatted.
The limiting reagant is the # of white hats. Person A has the largest single bit of information. Let's start with what he sees.
If both Person B and C had white hats, Person A would know he's black-hatted. Therefore, Person B & C either have different color hats or are both black-hatted.
If Person B sees Person C with a white hat, then Person B knows he's black hatted, since Person B & C can't both have white hats according to Person A.
Person C knows that he can't be white-hatted if Person B is white-hatted or Person A would have different observations. Person C also knows that he can't be white-hatted if Person B is black-hatted, or Person B would know what's up. Therefore, Person C can only be black-hatted.
Non Homogeneous Rope Burning[tag: math, logic] - 11+ minutes, no idea yet^^, got as far as figuring out 30 minutes.. how can I get a 15 minutes block:D^^ + Show Spoiler +You have two ropes, each of which takes one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using these non-homogeneous ropes and a lighter, time 45 minutes. Note: Some clarification on what is meant by non-homogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaining 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off. Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox] + Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Footsize and Spelling Ability [tag: thinkoutsidethebox] + Show Spoiler +Scientific studies have shown that there is a direct, positive correlation between foot size and performance in spelling bees / spelling tests. How can you explain this correlation? Cork, Bottle, Coin [tag: thinkoutsidethebox] + Show Spoiler +If you were to put a coin into an empty bottle and then insert a cork in the bottle's opening, how could you remove the coin without taking out the cork or breaking the bottle? Glass Half Full [tag: math] + Show Spoiler +You are in an empty room and you have a transparent glass of water. The glass is a right cylinder, and it looks like it's half full, but you're not sure. How can you accurately figure out whether the glass is half full, more than half full, or less than half full? You have no ruler or writing utensils. Hint: + Show Spoiler + tagged math because geometry is math. Faustian Round Table Game [tag: logic] + Show Spoiler +You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a round table. He gives himself and you a huge pile of quarters. He says "ok, we'll take turns putting quarters down, no overlapping allowed, and the quarters must rest on the table surface. the first guy who can't put a quarter down loses." you guys are about to start playing, and the devil says that he'll go first. however, at this point you immediately interject, and ask if you can go first instead. you make this interjection because you are very smart, and you know that if you go first, you can guarantee victory. explain how you can guarantee victory. Note: If you can put a quarter down so that it balances on the edge, then so can the devil Red Eyes and Blue Eyes [tag: logic] + Show Spoiler +There is an island of monks where everyone has either brown eyes or red eyes. Monks who have red eyes are cursed, and are supposed to commit suicide at midnight. However, no one ever talks about what color eyes they have, because the monks have a vow of silence. Also, there are no reflective surfaces on the whole island. Thus, no one knows their own eye color; they can only see the eye colors of other people, and not say anything about them. Life goes on, with brown-eyed monks and red-eyed monks living happily together in peace, and no one ever committing suicide. Then one day a tourist visits the island monastery, and, not knowing that he's not supposed to talk about eyes, he states the observation "At least one of you has red eyes." Having acquired this new information, something dramatic happens among the monks. What happens? Where's the father? [tag: math, thinkoutsidethebox] + Show Spoiler +The mother is 21 years older than the child. In 6 years from now, the mother will be 5 times as old as the child. Question: Where's the father? Three lightbulbs[tag: thinkoutsidethebox] + Show Spoiler +You are in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You are allowed only one chance to enter the room with the light bulbs. How can you determine which lightswitch goes with which light bulb? Manholes + Show Spoiler + Forcefield Detainment [tag: logic] + Show Spoiler +A group of prisoners are trapped in a forcefield. These prisoners are perfectly brave, meaning that they would attempt an escape on any positive probability of success. The prisoners are monitored by a guard who has only one bullet in his gun, but who also has perfect marksmanship skills (he never misses). A maintenance technician needs to tune up the forcefield generator, and so for one second, the forcefield is released. How can the guard still keep all the prisoners detained? Note: The prisoners are amoral, so they will back out of agreements with each other made a priori if they are sure they will be the one dying. Also, assume no prisoner can be manhandled out of the forcefield radius by the other prisoners. Lemming Drownings[tag: logic] + Show Spoiler +Somewhere in Northern Eurasia, a group of 20 lemmings is planning a special group suicide this year. Each of the lemmings will be placed in a random position along a thin, 100 meter long plank of wood which is floating in the sea. Each lemming is equally likely to be facing either end of the plank. At time t=0, all the lemmings walk forward at a slow speed of 1 meter per minute. If a lemming bumps into another lemming, the two both reverse directions. If a lemming falls off the plank, he drowns. What is the longest time that must elapse till all the lemmings have drowned? Hint: + Show Spoiler +After making a certain observation, you'll find the calculations trivial Infinite Quarters [tag: math, logic] + Show Spoiler +You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all? 12 balls [tag: math, logic] + Show Spoiler +You have 12 identical-looking balls. One of these balls has a different weight from all the others. You also have a two-pan balance for comparing weights. Using the balance only 3 times, how can you determine which ball has the unique weight, and also determine whether it is heavier or lighter than the others? Coin Machine Weighing [tag: math, logic] + Show Spoiler +You have 12 identical-looking balls. One of these balls has a different weight from all the others. You also have a two-pan balance for comparing weights. Using the balance in the smallest number of times possible, determine which ball has the unique weight, and also determine whether it is heavier or lighter than the others.
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infinite quarters: flip all quarters. split pile anywhere. cardinality is the same: aleph null on both sides.
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United States1719 Posts
Hey, coin machine's spoiler contains the same question as the 12 balls question, could you change that?
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+ Show Spoiler +Circles are the only shape of a hole where a slightly larger cover is guaranteed not to fall into it. All other shapes, you can rotate the cover a certain way to make it fall into the hole.
Three lightbulbs[tag: thinkoutsidethebox] + Show Spoiler +You are in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You are allowed only one chance to enter the room with the light bulbs. How can you determine which lightswitch goes with which light bulb? + Show Spoiler +Leave one of them on for a long time, then turn it off and turn on another switch and enter the room. The one that is currently lit is obvious, then feel the two other lightbulbs, the one you left on for a long time will still be hot.
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Infinite Quarters [tag: math, logic]+ Show Spoiler + You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all?
+ Show Spoiler +Flip all the quarters over, then divide the table how ever you want. If the table and the amount of quarters are infinite then on either side of the split there should be an infinite amount of tails.
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On December 05 2010 15:49 Trion wrote:Infinite Quarters [tag: math, logic] + Show Spoiler + You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all?
+ Show Spoiler +Flip all the quarters over, then divide the table how ever you want. If the table and the amount of quarters are infinite then on either side of the split there should be an infinite amount of tails.
On December 05 2010 15:45 annul wrote: infinite quarters: flip all quarters. split pile anywhere. cardinality is the same: aleph null on both sides.
fixed: now there are only a very large, but finite number of quarters. Theres still a simple way to solve this.
Also fixed the coin machine riddle repeat.
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I voted no for answers in the poll. I would like to change that to a yes. The curiosity is killing me and Google does not work.
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[QUOTE]On December 05 2010 13:21 rotinegg wrote:
The apple truck and apple-eating squirrel + Show Spoiler +There is an apple truck that can hold up to 1000 apples at any given time. It needs to make a delivery of 4000 apples from town A to B, which is 900 miles apart. However, there is a squirrel living in the truck that will eat 1 apple per mile. for example, if you load 1000 apples and take 4 trips, you will end up with 4x100 = 400 apples left. Try to maximize the number of apples you can deliver. Hint: + Show Spoiler +You can leave apples on the road along the way, and pick them up whenever you want, assuming your truck has the capacity to hold them.
+ Show Spoiler +Load up 1000 apples. Drive 250 miles, and take out the remaining 750. Then drive back and do the same thing three more times. At that point, you would have lost 250x4=1000 apples, leaving you with 3000 apples. Then load up 1000 apples and drive 333 miles. Leave the remaining 667, and drive back and repeat two more times. At this point, you would have lost an additional 999 apples, leaving you with 2001. Load up 1000 apples and drive the remaining 900-250-333=317 miles. Go back and do it again with another 1000 apples (sadly there's no way to save that last apple, I suggest you go ahead and eat it since this is a lot of work to be doing). Along the way you would have lost 317x2=634 apples. 2000-634=1366 apples.
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Willywutang can drown himself.
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so which riddles havent been solved so far?
i dont like doing them unless i can be the first to answer them lol
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United States1719 Posts
On December 05 2010 16:01 travis wrote: so which riddles havent been solved so far?
i dont like doing them unless i can be the first to answer them lol two of mine haven't been solved, they're in the second reply to OP
edit: The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 The mars robots (Some computer science knowledge required) + Show Spoiler +NASA is trying to launch two robots onto mars so they can construct a station. This is a two robot job, so both robots must work together to build the station. However, they do not have a spaceship big enough to fit both, so decide to launch them separately. Although both rockets will be following the same launch protocols, some error will be inevitable and the robots will land in different positions. Now assume that mars is an endless array of trillions and zillions of squares, stretched side-by-side. Basically, it is an infinite one-dimensional space. Your goal is to program the same algorithm on both robots so when they land, they will start looking for each other, and construct the station together. The robots CANNOT be programmed with different algorithms, and have no means of communication to earth or each other or any entity whatsoever. However, the robots will deploy a parachute when landing, and that parachute will remain in the square they land forever. The robot can detect whether they have found a parachute in the square it is in. Come up with an algorithm so that they will be guaranteed to find each other. (Normal restrictions in programming apply, such as integer overflow: any number larger than 2^32 will be undefined, so if you are keeping a counter and it reaches above that, it will render your algorithm useless.)
quoted per your convenience
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On December 05 2010 16:01 travis wrote: so which riddles havent been solved so far?
i dont like doing them unless i can be the first to answer them lol
Three lightbulbs[tag: thinkoutsidethebox]
Forcefield Detainment [tag: logic]
Lemming Drownings[tag: logic]
Infinite Quarters [tag: math, logic] (now finite) + Show Spoiler +
12 balls [tag: math, logic]
The blind man and checker board
The mars robots (Some computer science knowledge required)
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Coin Machine Weighing [tag: math, logic]+ Show Spoiler + you have 20 coin machines, each of which produce the same kind of coin. you know how much a coin is supposed to weigh. one of the machines is defective, in that every coin it produces weighs 1 ounce less than it is supposed to. you also have an electronic weighing machine. how can you determine which of the 20 machines is defective with only one weighing? (by one use, we mean you put a bunch of stuff on the machine and read a number, and that's it -- you not allowed to accumulate weight onto the machine and watch the numbers ascend, because that's just like multiple weighings). you are allowed to crank out as many coins from each machine as you like.
+ Show Spoiler +1.label each machine 1-20. 2.Get one coin for the first machine 2 from the second 3 from the third ect. 3.Weigh them all 4. If it is one ounce to light then it is machine one, if it is 2 ounces to light it is machine to, if it is 3 ounces to light it the third machine, ect.
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Infinite Quarters [tag: math, logic] - Hide Spoiler -
[/spoiler]You are wearing a blindfold and thick gloves. An infinite number of quarters are laid out before you on a table of infinite area. Someone tells you that 20 of these quarters are tails and the rest are heads. He says that if you can split the quarters into 2 piles where the number of tails quarters is the same in both piles, then you win all of the quarters. You are allowed to move the quarters and to flip them over, but you can never tell what state a quarter is currently in (the blindfold prevents you from seeing, and the gloves prevent you from feeling which side is heads or tails). How do you partition the quarters so that you can win them all?[/spoiler]
+ Show Spoiler +Just grab 20 quarters and flip them over into the 2nd pile. Even if you grab some tails quarters you will be flipping them to heads so they are removed from both sides.
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For Forcefield Detainment is there a rule about what prisoner the guard shoots at?
+ Show Spoiler + Because if he always shoots at the one that moves toward escaping first, then he doesn't have to do anything, none of them will move.
I guess my answer is: + Show Spoiler + The guard has to say "I will shoot at the first person who moves when the force field is deactivated." But then that might not work because the prisoners might not believe him, and that gives them a positive probability of escaping.
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added answers. The ones I didn't add answers for have answers somewhere in this thread, they are just really long and I didn't feel like typing them out. The exception is the 12 balls one where I googled the answer because it takes such a freaking long logical explanation.
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The blind man and checker board + Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
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On December 05 2010 16:02 rotinegg wrote:Show nested quote +On December 05 2010 16:01 travis wrote: so which riddles havent been solved so far?
i dont like doing them unless i can be the first to answer them lol two of mine haven't been solved, they're in the second reply to OP edit: Show nested quote +The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 The mars robots (Some computer science knowledge required) + Show Spoiler +NASA is trying to launch two robots onto mars so they can construct a station. This is a two robot job, so both robots must work together to build the station. However, they do not have a spaceship big enough to fit both, so decide to launch them separately. Although both rockets will be following the same launch protocols, some error will be inevitable and the robots will land in different positions. Now assume that mars is an endless array of trillions and zillions of squares, stretched side-by-side. Basically, it is an infinite one-dimensional space. Your goal is to program the same algorithm on both robots so when they land, they will start looking for each other, and construct the station together. The robots CANNOT be programmed with different algorithms, and have no means of communication to earth or each other or any entity whatsoever. However, the robots will deploy a parachute when landing, and that parachute will remain in the square they land forever. The robot can detect whether they have found a parachute in the square it is in. Come up with an algorithm so that they will be guaranteed to find each other. (Normal restrictions in programming apply, such as integer overflow: any number larger than 2^32 will be undefined, so if you are keeping a counter and it reaches above that, it will render your algorithm useless.) quoted per your convenience
i fear that since it hasnt been yet solved ill be wrong, but checkerboard part 1
+ Show Spoiler + 1,1,1,1 2,1,1,1 1,2,1,1 1,1,2,1 1,1,1,2 2,2,1,1 2,1,1,2 1,2,1,2
8 turns counting a turn where he flips no coins or flips all of them
i flat out don't understand how part 2 would be solved
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On December 05 2010 16:22 ZapRoffo wrote:For Forcefield Detainment is there a rule about what prisoner the guard shoots at? + Show Spoiler + Because if he always shoots at the one that moves toward escaping first, then he doesn't have to do anything, none of them will move.
I guess my answer is: + Show Spoiler + The guard has to say "I will shoot at the first person who moves when the force field is deactivated." But then that might not work because the prisoners might not believe him, and that gives them a positive probability of escaping.
Assume they can collaborate to all move at the exact same time, or the lock hands and swing around such that if they let go when the field is turned off it is effectively out of their control who leaves the forcefield first. Then your solution doesn't work. You're on the right track though. Just takes a little more logic.
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double post my bad, meant to edit not quote.
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3 lightbulbs:
You are in a room with three light switches, each of which controls one of three light bulbs in the next room. Your task is to determine which switch controls which bulb. All lights are initially off, and you can't see into one room from the other. You are allowed only one chance to enter the room with the light bulbs. How can you determine which lightswitch goes with which light bulb?
answer:
+ Show Spoiler + flip switch 1 and switch 2. leave them on for say, 10 minutes. flip switch 2 back the other way. go into the room. feel both light bulbs that are off. the warm one is the switch u turned on and then off.
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On December 05 2010 16:33 t3tsubo wrote:Show nested quote +On December 05 2010 16:22 ZapRoffo wrote:For Forcefield Detainment is there a rule about what prisoner the guard shoots at? + Show Spoiler + Because if he always shoots at the one that moves toward escaping first, then he doesn't have to do anything, none of them will move.
I guess my answer is: + Show Spoiler + The guard has to say "I will shoot at the first person who moves when the force field is deactivated." But then that might not work because the prisoners might not believe him, and that gives them a positive probability of escaping.
Assume they can collaborate to all move at the exact same time, or the lock hands and swing around such that if they let go when the field is turned off it is effectively out of their control who leaves the forcefield first. Then your solution doesn't work. You're on the right track though. Just takes a little more logic.
+ Show Spoiler +Claim that in x amount of seconds after the force filed goes down the farthest person away from it will be shot.
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Umm for the round table one, you could just put all the quarters on the table, because it was not stated that you had to do only one at a time ^_^
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For the 12 balls one that awnser is stupid and done the wrong way, put 6 and 6 one will be heavier, from the heavier group do 3 and 3, and from the heavier group of that one measure 1 on each side if both are equal the one you didnt measure is heavier if 1 is heavier then the other then thats the heavier ball.
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Non Homogeneous Rope Burning[tag: math, logic] + Show Spoiler +You have two ropes, each of which takes one hour to burn completely. Both of these ropes are non-homogeneous in thickness, meaning that some parts of the ropes are chunkier than other parts of the rope. using these non-homogeneous ropes and a lighter, time 45 minutes. Note: Some clarification on what is meant by non-homogeneous. For instance, maybe a particular section of rope that is 1/8 of the total length is really chunky, and takes 50 minutes to burn off. then it would take 10 minutes to burn off the remaining 7/8, since we know that the whole rope takes an hour to burn off. that's just an example; we don't know any such ratios beforehand. The point is, if you look at one of your ropes and cut it into pieces, you have no clue how long any individual piece will take to burn off. + Show Spoiler [Solution] + Just burn rope N1 from both ends and rope N2 from one end at the same time. When rope N1 burns completely you have 30 minutes measured (it doesnt matter about the distribution of the mass in the rope since both ends meet when the time left to burn is the same for each end). Then at that right 30 minutes you start fire on the second end of rope N2 (since 30 min passed there are 30 minutes left to burn in rope N2, wich burning now from both ends will mesure 15 min). When rope N2 burns 45 minutes is measured.
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United States4053 Posts
On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler [part 1 answer] +7. 1000 0100 0010 0001 0101 0110 0011 as for part 2 + Show Spoiler + I seem to remember that you simply perform a set of moves in order and it works regardless of how the table turns
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mars rover: + Show Spoiler + With my limited computer science knowlage: Have both rover move in spirals that continualy getting larger (i.e. after the begining move X, turn 90, move X, turn 90, move X, turn 90, move X, turn 90, move X+1...) end subsequence when parashute is found. Move robot to midpoint of start position and current location, begin construction. The rest of the solution is just symatics. In the event the midpoint is on a decimal, round up if location is negative and down if positive. load in a sequence to make sure the roaver will not fall of cliffs. If the robots need to start at the same time, then have the robot calculate the position of the other when the parishute is found (just 2*X,2*y) and calculate how long that roaver will take to get to the construction site. Thats all the smantics I can think of, am I right?
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On December 05 2010 16:02 rotinegg wrote:Show nested quote +On December 05 2010 16:01 travis wrote: so which riddles havent been solved so far?
i dont like doing them unless i can be the first to answer them lol two of mine haven't been solved, they're in the second reply to OP edit: Show nested quote +The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 The mars robots (Some computer science knowledge required) + Show Spoiler +NASA is trying to launch two robots onto mars so they can construct a station. This is a two robot job, so both robots must work together to build the station. However, they do not have a spaceship big enough to fit both, so decide to launch them separately. Although both rockets will be following the same launch protocols, some error will be inevitable and the robots will land in different positions. Now assume that mars is an endless array of trillions and zillions of squares, stretched side-by-side. Basically, it is an infinite one-dimensional space. Your goal is to program the same algorithm on both robots so when they land, they will start looking for each other, and construct the station together. The robots CANNOT be programmed with different algorithms, and have no means of communication to earth or each other or any entity whatsoever. However, the robots will deploy a parachute when landing, and that parachute will remain in the square they land forever. The robot can detect whether they have found a parachute in the square it is in. Come up with an algorithm so that they will be guaranteed to find each other. (Normal restrictions in programming apply, such as integer overflow: any number larger than 2^32 will be undefined, so if you are keeping a counter and it reaches above that, it will render your algorithm useless.) quoted per your convenience
Mars robots:
+ Show Spoiler + When the robots land, they will either start moving towards the sun if it's up, or wait until it's up if it's down. This initial condition is only checked once, and determines which direction the robots move in. The robots move at half speed. Because they are both moving in the same direction, one of them is bound to approach the parachute of the other, at which point it ramps up to full speed. It will overtake the other robot in the same amount of time it took to reach the parachute.
Previous poster - Mars is one dimensional in this puzzle (infinitely straight line.) A simple issue with your solution is - what happens when the two rovers land further apart than it is possible to represent in whatever format you choose? At some point you won't be able to spiral bigger.
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^^ the surface of mars is finite, and if each point in our matrix is one square foot, then it does not violate the 2^32 rule
edit: aww damn, I just read the original problem better.
On December 05 2010 16:33 t3tsubo wrote:Show nested quote +On December 05 2010 16:22 ZapRoffo wrote:For Forcefield Detainment is there a rule about what prisoner the guard shoots at? + Show Spoiler + Because if he always shoots at the one that moves toward escaping first, then he doesn't have to do anything, none of them will move.
I guess my answer is: + Show Spoiler + The guard has to say "I will shoot at the first person who moves when the force field is deactivated." But then that might not work because the prisoners might not believe him, and that gives them a positive probability of escaping.
Assume they can collaborate to all move at the exact same time, or the lock hands and swing around such that if they let go when the field is turned off it is effectively out of their control who leaves the forcefield first. Then your solution doesn't work. You're on the right track though. Just takes a little more logic. + Show Spoiler +The guard anounces that he will shoot the tallest escapee (or any other quantifyable trait) The tallest will have 100% odds of being killed, so won't try, then the second won't try, and the shortest is alone, and won't try.
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probability (quite a well known one):
+ Show Spoiler +Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
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On December 05 2010 18:18 Ghazwan wrote:probability (quite a well known one): + Show Spoiler +Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? also known as Monty Hall paradox
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On December 05 2010 18:20 Geo.Rion wrote:Show nested quote +On December 05 2010 18:18 Ghazwan wrote:probability (quite a well known one): + Show Spoiler +Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? also known as Monty Hall paradox
yup
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On December 05 2010 17:53 infinitestory wrote:Show nested quote +On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler [part 1 answer] +7. 1000 0100 0010 0001 0101 0110 0011 as for part 2 + Show Spoiler + I seem to remember that you simply perform a set of moves in order and it works regardless of how the table turns
How are you transitioning from one permutation to the next? You can only flip one coin at a time. What you're doing is flipping two coins at a time, in some cases.
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On December 05 2010 18:23 jeeneeus wrote:Show nested quote +On December 05 2010 17:53 infinitestory wrote:On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler [part 1 answer] +7. 1000 0100 0010 0001 0101 0110 0011 as for part 2 + Show Spoiler + I seem to remember that you simply perform a set of moves in order and it works regardless of how the table turns
How are you transitioning from one permutation to the next? You can only flip one coin at a time. What you're doing is flipping two coins at a time, in some cases.
It says you can flip as many coins as you want
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On December 05 2010 17:10 SubtleSense wrote: For the 12 balls one that awnser is stupid and done the wrong way, put 6 and 6 one will be heavier, from the heavier group do 3 and 3, and from the heavier group of that one measure 1 on each side if both are equal the one you didnt measure is heavier if 1 is heavier then the other then thats the heavier ball.
If you measure the heavier set of 6 and the ball is supposed to be lighter then you would be wrong (because you need to find out which ball is the off-weighted one). I don't quite know how to answer this one. I would almost guess that it's something like measure 4 balls on one side and 8 on the other but that doesn't work either.
On December 05 2010 18:18 Ghazwan wrote:probability (quite a well known one): + Show Spoiler +Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?
statistically you have a 2/3rd's chance of winning if you switch to door 2- This is correct But logically you would make a new equation and have 1/2 chance if you switch.
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FORCEFIELD DETAINMENT + Show Spoiler + The guard announces that he will shoot whoever is to the leftmost side of him trying to escape. Rather self-explanatory. Basically the same as the 'foremost person being shot' but does not allow for handholding or other strategies. Even if they swing around and let go, the leftmost person will still back out of the agreement as he does not wish to be shot. The next people will then back out and the chain effect will commence.
Oh and I love the Monty Hall paradox, existing probability is transferred to the unknown variable. This logic also transfers to the 3 prisoners paradox. This changed my outlook on probability altogether (that the probability you perceive is based on your perspective. After all, if there were an omnipresent being, he would know everything and the answer would already be clear to him), and has become one of my favorite areas of mathematics.
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On December 05 2010 17:10 SubtleSense wrote: For the 12 balls one that awnser is stupid and done the wrong way, put 6 and 6 one will be heavier, from the heavier group do 3 and 3, and from the heavier group of that one measure 1 on each side if both are equal the one you didnt measure is heavier if 1 is heavier then the other then thats the heavier ball.
your solution doesn't work because it is not specified whether the ball is lighter or heavier than the others at the beginning and you're assuming it has to be heavier.
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United States1719 Posts
Monty Hall paradox:Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? + Show Spoiler +It is in your advantage to switch your choice, as he is essentially giving you a package of two doors, only he already eliminated the one that doesn't have the prize for you. So you will have a 2/3 chance to win if you switch. If that's hard to understand, you can extrapolate it to 100 doors. You would definitely switch if he eliminated 98 doors for you and asked you if you wanted to switch to the other available door.
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For the manhole one, I think you mean manhole covers.
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On December 05 2010 13:38 wishbones wrote: Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I? + Show Spoiler +
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On December 06 2010 03:24 beefhamburger wrote: For the manhole one, I think you mean manhole covers. It's a riddle made famous as a microsoft interview question, so I kept the original wording. Aka interviewees for a job and microsoft would get asked this question and their responses judged.
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On December 05 2010 13:38 wishbones wrote: Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I?
The uncle of that man.
+ Show Spoiler +That mans father, is my father's son, so the father of that man is is the son of your father, so that mans father is your brother, and therefore you are that mans uncle.
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On December 05 2010 19:23 ixi.genocide wrote:Show nested quote +On December 05 2010 17:10 SubtleSense wrote: For the 12 balls one that awnser is stupid and done the wrong way, put 6 and 6 one will be heavier, from the heavier group do 3 and 3, and from the heavier group of that one measure 1 on each side if both are equal the one you didnt measure is heavier if 1 is heavier then the other then thats the heavier ball. If you measure the heavier set of 6 and the ball is supposed to be lighter then you would be wrong (because you need to find out which ball is the off-weighted one). I don't quite know how to answer this one. I would almost guess that it's something like measure 4 balls on one side and 8 on the other but that doesn't work either. Show nested quote +On December 05 2010 18:18 Ghazwan wrote:probability (quite a well known one): + Show Spoiler +Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which he knows has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice? statistically you have a 2/3rd's chance of winning if you switch to door 2- This is correct But logically you would make a new equation and have 1/2 chance if you switch.
how is that logical? you have 66% chance to win if you switch, not 50%. statistically, logically, however you wanna call it
the simplest way i found to explain it is: you only lose by switching if you picked right to begin with. what's the chance of you picking right to begin with? boom.
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On December 06 2010 04:35 Trion wrote:Show nested quote +On December 05 2010 13:38 wishbones wrote: Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I? The uncle of that man. + Show Spoiler +That mans father, is my father's son, so the father of that man is is the son of your father, so that mans father is your brother, and therefore you are that mans uncle.
You dont have any brother or sisters - how can you be an uncle.....
You are pointing to your own child.
That mans father (= your kid's father = yourself) is my fathers son (= the son of your father = yourself - no sibs remember)
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Heres a riddle that i enjoyed quite a bit, because it got me thinking for quite some time (perhaps one of you could be a lot faster than I was). The story feels kinda .. REALLY constructed, but i like it anyway Imagine a dwarf kingdom with two different groups of dwarfs. In one group of dwarfs, everyone is wearing a blue hat/helm, in the other group, every dwarf is wearing a red one. In order to honor their king, they have to present themselves in a line, with all red dwarfs on the right side, and all blue dwarfs on the left side. The dwarfs are not allowed to communicate with each other in any way possible (no talking, no phantomime, no silly dancing, nothing). Also, although the dwarfs can obviously see the other dwarfs hats, they do not know which color their own hat has (and are not allowed to ask the other dwarfs, as stated previously).
Question : How do the dwarfs manage to line up like they are supposed to do ?
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Maybe his brother is dead.
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On December 06 2010 04:38 Sanguinarius wrote:Show nested quote +On December 06 2010 04:35 Trion wrote:On December 05 2010 13:38 wishbones wrote: Brothers and Sisters, I have none. But that man's father is my father's son.
WHO AM I? The uncle of that man. + Show Spoiler +That mans father, is my father's son, so the father of that man is is the son of your father, so that mans father is your brother, and therefore you are that mans uncle.
You dont have any brother or sisters - how can you be an uncle..... You are pointing to your own child. That mans father (= your kid's father = yourself) is my fathers son (= the son of your father = yourself - no sibs remember) That makes sense. My mistake. I missed the first bit.
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Nice thread, only remark I can make about it is that a lot of these are well-known.
Monty Hall is nice but also very clear-cut. I'd still like a good explanation (in layman's terms) for this phenomenon, which is closely related:
Say you're given two envelopes, each one with a real amount of money in it. You have no clue how much is in each, but the only information that's given about it is that one contains double the amount of the other. So you have to pick one and take that amount of money.
You can pick either, open it, look at the amount, then switch and pick the other. The paradox is this: Just picking one (which would give you amount X) is not as good as picking one first, then switching to the other envelope, which gives you (0.5 * X * 0.5 + 0.5 * X * 2) / 2, which is 1.25 * X. But that means you could just divert from your initial choice and get more? Please explain!
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On December 06 2010 04:40 Espelz wrote:Heres a riddle that i enjoyed quite a bit, because it got me thinking for quite some time (perhaps one of you could be a lot faster than I was). The story feels kinda .. REALLY constructed, but i like it anyway Imagine a dwarf kingdom with two different groups of dwarfs. In one group of dwarfs, everyone is wearing a blue hat/helm, in the other group, every dwarf is wearing a red one. In order to honor their king, they have to present themselves in a line, with all red dwarfs on the right side, and all blue dwarfs on the left side. The dwarfs are not allowed to communicate with each other in any way possible (no talking, no phantomime, no silly dancing, nothing). Also, although the dwarfs can obviously see the other dwarfs hats, they do not know which color their own hat has (and are not allowed to ask the other dwarfs, as stated previously). Question : How do the dwarfs manage to line up like they are supposed to do ?
hmm just intuitively, maybe i'm totally missing something, but + Show Spoiler + we want a BBB....BBRR....RR lineup in the end, right? so dwarf1 goes up and stands in line dwarf2 goes up and if dwarf1 is blue, he stands on the right. if red, on the left. so we'd have one of BB or BR or RR and.. just repeat the process? if BR, go in the middle so its either BRR or BBR and repeat
i don't really see the problem, lol probably missed something
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Math riddles are really dumb. This thread would be better with proper riddles, not generic collegial bonus math questions your professors gave you in your intermediate algebra classes.
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On December 06 2010 04:49JeeJee wrote:Show nested quote +On December 06 2010 04:40 Espelz wrote:Heres a riddle that i enjoyed quite a bit, because it got me thinking for quite some time (perhaps one of you could be a lot faster than I was). The story feels kinda .. REALLY constructed, but i like it anyway Imagine a dwarf kingdom with two different groups of dwarfs. In one group of dwarfs, everyone is wearing a blue hat/helm, in the other group, every dwarf is wearing a red one. In order to honor their king, they have to present themselves in a line, with all red dwarfs on the right side, and all blue dwarfs on the left side. The dwarfs are not allowed to communicate with each other in any way possible (no talking, no phantomime, no silly dancing, nothing). Also, although the dwarfs can obviously see the other dwarfs hats, they do not know which color their own hat has (and are not allowed to ask the other dwarfs, as stated previously). Question : How do the dwarfs manage to line up like they are supposed to do ? hmm just intuitively, maybe i'm totally missing something, but + Show Spoiler + we want a BBB....BBRR....RR lineup in the end, right? so dwarf1 goes up and stands in line dwarf2 goes up and if dwarf1 is blue, he stands on the right. if red, on the left. so we'd have one of BB or BR or RR and.. just repeat the process? if BR, go in the middle so its either BRR or BBR and repeat
i don't really see the problem, lol probably missed something
Actually you are right, it really is that simple... i just tend to think in the most complicated way possible, no matter what problem Therefore it took me kinda long.
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On December 06 2010 05:02 CrY. wrote: Math riddles are really dumb. This thread would be better with proper riddles, not generic collegial bonus math questions your professors gave you in your intermediate algebra classes. I generally agree, thats why most of the ones in my OP don't have much to do with calculations, except basic ones.
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[QUOTE]On December 05 2010 13:21 rotinegg wrote: I love riddles so much Solutions to some of them are below. Meanwhile, here's some riddles of my own that I picked up over the years!
The apple truck and apple-eating squirrel + Show Spoiler +There is an apple truck that can hold up to 1000 apples at any given time. It needs to make a delivery of 4000 apples from town A to B, which is 900 miles apart. However, there is a squirrel living in the truck that will eat 1 apple per mile. for example, if you load 1000 apples and take 4 trips, you will end up with 4x100 = 400 apples left. Try to maximize the number of apples you can deliver. Hint: + Show Spoiler +You can leave apples on the road along the way, and pick them up whenever you want, assuming your truck has the capacity to hold them.
Answer + Show Spoiler +The goal is to have 2000 total at whatever distance you can get them. So to get 2000 total you load up 1000 4x and drop them off at 500 miles. This means you have 2000 apples at 400 miles away. Two trips of 1000 apples gives you 1200 apples at your final destination.
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United States1719 Posts
On December 06 2010 05:35 insectoceanx wrote:Show nested quote +On December 05 2010 13:21 rotinegg wrote:I love riddles so much Solutions to some of them are below. Meanwhile, here's some riddles of my own that I picked up over the years! The apple truck and apple-eating squirrel+ Show Spoiler +There is an apple truck that can hold up to 1000 apples at any given time. It needs to make a delivery of 4000 apples from town A to B, which is 900 miles apart. However, there is a squirrel living in the truck that will eat 1 apple per mile. for example, if you load 1000 apples and take 4 trips, you will end up with 4x100 = 400 apples left. Try to maximize the number of apples you can deliver. Hint: + Show Spoiler +You can leave apples on the road along the way, and pick them up whenever you want, assuming your truck has the capacity to hold them. Answer + Show Spoiler +The goal is to have 2000 total at whatever distance you can get them. So to get 2000 total you load up 1000 4x and drop them off at 500 miles. This means you have 2000 apples at 400 miles away. Two trips of 1000 apples gives you 1200 apples at your final destination. you can do better
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Here's one that is not to hard.
The square house: + Show Spoiler +A man builds a square house and puts a window on each side of the house. All the windows are facing south. One day the man sees a bear outside the window, what colour is the bear? Answer:+ Show Spoiler +Each window is facing south means that the house is on the north pole, the bear is a polar bear, which of course is white.
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United States4053 Posts
On December 06 2010 05:41 rotinegg wrote:Show nested quote +On December 06 2010 05:35 insectoceanx wrote:On December 05 2010 13:21 rotinegg wrote:I love riddles so much Solutions to some of them are below. Meanwhile, here's some riddles of my own that I picked up over the years! The apple truck and apple-eating squirrel+ Show Spoiler +There is an apple truck that can hold up to 1000 apples at any given time. It needs to make a delivery of 4000 apples from town A to B, which is 900 miles apart. However, there is a squirrel living in the truck that will eat 1 apple per mile. for example, if you load 1000 apples and take 4 trips, you will end up with 4x100 = 400 apples left. Try to maximize the number of apples you can deliver. Hint: + Show Spoiler +You can leave apples on the road along the way, and pick them up whenever you want, assuming your truck has the capacity to hold them. Answer + Show Spoiler +The goal is to have 2000 total at whatever distance you can get them. So to get 2000 total you load up 1000 4x and drop them off at 500 miles. This means you have 2000 apples at 400 miles away. Two trips of 1000 apples gives you 1200 apples at your final destination. you can do better + Show Spoiler + Take 4 trips of 1000 for 250 miles each to get 3000 at 250 miles. Then take 3 trips of 1000 for 1000/3 miles to get 2000 at 1750/3 miles. Drive these the rest of the way to get 4100/3 apples, which is about 1367
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I really like this one:
Three Princesses + Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
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On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
You obviously want to marry the oldest daughter, but when the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you will know which daughter is oldest?
why would you want to marry the oldest one? <,< depends on ages i suppose, but if it's like 10 20 40 i'd vote for 20y/o on that note, + Show Spoiler +if i were to ask you "are you the middle daughter?" would you say yes?" the young daughter would say 'yes' to the middle question so 'no' overall. the oldest daughter would say 'no' to the middle question so 'no' overall and the middle daughter (lying) would say 'no' to the middle question so 'yes' overall and the middle daughter (truth) would say 'yes' to the middle question so 'yes' overall aka marry the one that says yes
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United States4053 Posts
On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
You obviously want to marry the oldest daughter, but when the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you will know which daughter is oldest?
+ Show Spoiler [not an answer] +ugh... this one hurts my head... seems to me that if you ask the middle daughter you're screwed... @above: I think the middle daughter randomly picks whether to lie or say the truth... like she could lie for the middle question but then answer truthfully overall
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On December 06 2010 06:33 JeeJee wrote:Show nested quote +On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
You obviously want to marry the oldest daughter, but when the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you will know which daughter is oldest?
why would you want to marry the oldest one? <,< depends on ages i suppose, but if it's like 10 20 40 i'd vote for 20y/o on that note, + Show Spoiler +if i were to ask you "are you the middle daughter?" would you say yes?" the young daughter would say 'yes' to the middle question so 'no' overall. the oldest daughter would say 'no' to the middle question so 'no' overall and the middle daughter (lying) would say 'no' to the middle question so 'yes' overall and the middle daughter (truth) would say 'yes' to the middle question so 'yes' overall aka marry the one that says yes
I'm sorry, i had forgotten some of the riddle, updated it now
Also, on your answer:
+ Show Spoiler +You can't ask them all one question; you get one question and you need to choose one daughter to ask it to.
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On December 06 2010 06:34 DumEN wrote:Show nested quote +On December 06 2010 06:33 JeeJee wrote:On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
You obviously want to marry the oldest daughter, but when the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you will know which daughter is oldest?
why would you want to marry the oldest one? <,< depends on ages i suppose, but if it's like 10 20 40 i'd vote for 20y/o on that note, + Show Spoiler +if i were to ask you "are you the middle daughter?" would you say yes?" the young daughter would say 'yes' to the middle question so 'no' overall. the oldest daughter would say 'no' to the middle question so 'no' overall and the middle daughter (lying) would say 'no' to the middle question so 'yes' overall and the middle daughter (truth) would say 'yes' to the middle question so 'yes' overall aka marry the one that says yes I'm sorry, i had forgotten some of the riddle, updated it now Also, on your answer: + Show Spoiler +You can't ask them all one question; you get one question and you need to choose one daughter to ask it to.
you just totally changed your question lol fine, then marry the one that says no or the one that doesn't say yes to my question, only one question necessary
edit:
On December 06 2010 06:34 infinitestory wrote:Show nested quote +On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
You obviously want to marry the oldest daughter, but when the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you will know which daughter is oldest?
+ Show Spoiler [not an answer] +ugh... this one hurts my head... seems to me that if you ask the middle daughter you're screwed... @above: I think the middle daughter randomly picks whether to lie or say the truth... like she could lie for the middle question but then answer truthfully overall
that could be a problem. edit: + Show Spoiler +but i don't think so actually, i remembered why i figured this one out fairly quickly.. i actually posted basically the same question before, here:http://www.teamliquid.net/blogs/viewblog.php?id=112187
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Yeh, sorry bout that =/
+ Show Spoiler + They only answer to the final question, which is "would you say yes?"
If the daughter you asked your question to were the youngest, she would answer yes. If the daughter was the oldest, she would answer no. If it was the middle one it would be either yes or no.
So you get nothing from asking that question, the one you asked could still be anyone.
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On December 06 2010 06:43 DumEN wrote:Yeh, sorry bout that =/ + Show Spoiler + They only answer to the final question, which is "would you say yes?"
If the daughter you asked your question to were the youngest, she would answer yes. If the daughter was the oldest, she would answer no. If it was the middle one it would be either yes or no.
So you get nothing from asking that question, the one you asked could still be anyone.
+ Show Spoiler +i highly disagree.. of course they only answer the final question, but the final question's answer depends on what they would have answered to the other question
for example, if i come up to you and say "if i were to ask you, "is your nickname dumen on TL", would you say yes?" the thought process is "yes my nickname is dumen, so yes i'd say yes to that question --> yes
it works.
and to clarify, i don't see how the youngest daughter would say yes here's the question again if i were to ask you "are you the middle daughter?" would you say yes?" she's thinking *ok i'm not the middle daughter, so i'd lie to that question and answer 'yes', so i would say yes but i can't tell him that since i'm a liar so i say*: no
repeat same reasoning for the other 2 daughters
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Love the thread, huge fan of this stuff.
One of my favourites goes like this (didnt see it posted yet):
+ Show Spoiler +Every man in a village of fifty couples has been unfaithful to his wife. Every woman in the village instantly knows when a man other than her husband has philandered (you know how small towns are) but not when her own husband has ("always the last to know"). The village's notolerance adultery statute requires that a woman who can prove her husband is unfaithful must kill him that very day. No woman would dream of disobeying this law.
One day, the queen, who is known to be infallible, visits the village. She announces that at least one husband has been unfaithful.
What happens?
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You're in a room, with no windows, no doors, no nothing. Only a mirror. Question 1: How did you get in there? Question 2: How do you get out?
Removing the answer for a bit =)
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On December 06 2010 06:48 JeeJee wrote:Show nested quote +On December 06 2010 06:43 DumEN wrote:Yeh, sorry bout that =/ + Show Spoiler + They only answer to the final question, which is "would you say yes?"
If the daughter you asked your question to were the youngest, she would answer yes. If the daughter was the oldest, she would answer no. If it was the middle one it would be either yes or no.
So you get nothing from asking that question, the one you asked could still be anyone.
+ Show Spoiler +i highly disagree.. of course they only answer the final question, but the final question's answer depends on what they would have answered to the other question
for example, if i come up to you and say "if i were to ask you, "is your nickname dumen on TL", would you say yes?" the thought process is "yes my nickname is dumen, so yes i'd say yes to that question --> yes
it works.
and to clarify, i don't see how the youngest daughter would say yes here's the question again if i were to ask you "are you the middle daughter?" would you say yes?" she's thinking *ok i'm not the middle daughter, so i'd lie to that question and answer 'yes', so i would say yes but i can't tell him that since i'm a liar so i say*: no
repeat same reasoning for the other 2 daughters
+ Show Spoiler + Ah yeah i failed at reading comprehension at that question, however: If you asked one of the daughters that, and she answers: "no" then you dont know which one it is, because the middle daughter might say that aswell.
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t3tsubo, please make an answer to Three hats before I go mad!
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On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn.
I think the Answer is partly correct or at least not precise: + Show Spoiler +if you start a fire lets say just before the end B of the island, the wind will blow it towards B and it will end there, leaving non-burnable space that is safe for Willywutang. You answer speaks of two meeting flames, which is only possible if the fire started at B or the wind is not blowing from A to B but from B to A.
or maybe it's a mistranslation of mine as to the direction of the wind
Edit: For Three Hats:
+ Show Spoiler +if A had seen 2 white hats he had known *I have a black hat*. So A saw either a white and a black or two black hats. If B had seen a white hat he had known *I have a black hat* because he know the two possible combinations for A to say no. He saw a black hat so he couldn't know. now you know that C was wearing a black hat.
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On December 06 2010 06:55 DumEN wrote:Show nested quote +On December 06 2010 06:48 JeeJee wrote:On December 06 2010 06:43 DumEN wrote:Yeh, sorry bout that =/ + Show Spoiler + They only answer to the final question, which is "would you say yes?"
If the daughter you asked your question to were the youngest, she would answer yes. If the daughter was the oldest, she would answer no. If it was the middle one it would be either yes or no.
So you get nothing from asking that question, the one you asked could still be anyone.
+ Show Spoiler +i highly disagree.. of course they only answer the final question, but the final question's answer depends on what they would have answered to the other question
for example, if i come up to you and say "if i were to ask you, "is your nickname dumen on TL", would you say yes?" the thought process is "yes my nickname is dumen, so yes i'd say yes to that question --> yes
it works.
and to clarify, i don't see how the youngest daughter would say yes here's the question again if i were to ask you "are you the middle daughter?" would you say yes?" she's thinking *ok i'm not the middle daughter, so i'd lie to that question and answer 'yes', so i would say yes but i can't tell him that since i'm a liar so i say*: no
repeat same reasoning for the other 2 daughters
+ Show Spoiler + Ah yeah i failed at reading comprehension at that question, however: If you asked one of the daughters that, and she answers: "no" then you dont know which one it is, because the middle daughter might say that aswell.
+ Show Spoiler +middle daughter would say 'yes' regardless of whether she's lying or not unless you're saying that she randomly decides to say yes/no, in which case you can't answer this question
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United States4053 Posts
On December 06 2010 07:00 Slakkoo wrote: t3tsubo, please make an answer to Three hats before I go mad! + Show Spoiler + A has no idea what color his hat is. But the only way he would know is if B and C were both wearing white hats. So B and C are not both wearing white hats. Armed with this knowledge, B looks at C's hat. If he saw C had a white hat, B would know he had a black hat, since B and C are not both wearing white hats. Since B doesn't know the color of his hat either, C determines that he must be wearing a black hat.
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On December 06 2010 07:04 JeeJee wrote:Show nested quote +On December 06 2010 06:55 DumEN wrote:On December 06 2010 06:48 JeeJee wrote:On December 06 2010 06:43 DumEN wrote:Yeh, sorry bout that =/ + Show Spoiler + They only answer to the final question, which is "would you say yes?"
If the daughter you asked your question to were the youngest, she would answer yes. If the daughter was the oldest, she would answer no. If it was the middle one it would be either yes or no.
So you get nothing from asking that question, the one you asked could still be anyone.
+ Show Spoiler +i highly disagree.. of course they only answer the final question, but the final question's answer depends on what they would have answered to the other question
for example, if i come up to you and say "if i were to ask you, "is your nickname dumen on TL", would you say yes?" the thought process is "yes my nickname is dumen, so yes i'd say yes to that question --> yes
it works.
and to clarify, i don't see how the youngest daughter would say yes here's the question again if i were to ask you "are you the middle daughter?" would you say yes?" she's thinking *ok i'm not the middle daughter, so i'd lie to that question and answer 'yes', so i would say yes but i can't tell him that since i'm a liar so i say*: no
repeat same reasoning for the other 2 daughters
+ Show Spoiler + Ah yeah i failed at reading comprehension at that question, however: If you asked one of the daughters that, and she answers: "no" then you dont know which one it is, because the middle daughter might say that aswell.
+ Show Spoiler +middle daughter would say 'yes' regardless of whether she's lying or not unless you're saying that she randomly decides to say yes/no, in which case you can't answer this question
+ Show Spoiler +Yeah she says it randomly, and yes it can be solved. hint: + Show Spoiler +It is not a 'question in a question'.
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On December 06 2010 07:06 infinitestory wrote:Show nested quote +On December 06 2010 07:00 Slakkoo wrote: t3tsubo, please make an answer to Three hats before I go mad! + Show Spoiler + A has no idea what color his hat is. But the only way he would know is if B and C were both wearing white hats. So B and C are not both wearing white hats. Armed with this knowledge, B looks at C's hat. If he saw C had a white hat, B would know he had a black hat, since B and C are not both wearing white hats. Since B doesn't know the color of his hat either, C determines that he must be wearing a black hat.
Cheers!
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On December 06 2010 07:00 Slakkoo wrote: t3tsubo, please make an answer to Three hats before I go mad!
My answer, it is a pretty simple one: + Show Spoiler + * If A could see that B and C both had white hats (thus taking the only 2 white available) he would know that he had a black and say 'yes' when asked -> We know that B and C has at least 1 black hat (either one black and one white, or two black ones)
* B know that A didn't say 'yes' that means he knows that both he and C can not be wearing white hats. If he sees C having a white hat he knows that he himself must have a black one and would say answer 'yes' -> but he says 'no' as well and that give us that C must have a black one or B would have said 'yes'.
C answers that yes, now he knows he has a black one.
edit: damnit, not fast enough
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On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
+ Show Spoiler +No questions are necessary at first. you simply pick a princess at random and shoot her in the foot. If she says “ow” over and over, she’s the truthful one, if she says “This doesn’t hurt” over and over, she’s the liar, and if she alternates between the two answers, she’s the random one. if you choose the random one you simply choose to marry one of the other two and then proceed to ask your yes/no question which is “are you a princess?” if she says yes, she’s the honest one, and if she says no, she’s the liar, and this, my friends, is how it is done.
+ Show Spoiler +
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The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated).
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In regard to the four coins and the blind man and flipping as many coins as they want --
+ Show Spoiler +It is simple binary. The first solution showed this, but if you want to be specific and declare that you're only allowed to flip one time a round you would use gray code binary. http://en.wikipedia.org/wiki/Gray_code000 001 011 010 110 111 101 100
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On December 06 2010 06:49 xlat wrote:Love the thread, huge fan of this stuff. One of my favourites goes like this (didnt see it posted yet): + Show Spoiler +Every man in a village of fifty couples has been unfaithful to his wife. Every woman in the village instantly knows when a man other than her husband has philandered (you know how small towns are) but not when her own husband has ("always the last to know"). The village's notolerance adultery statute requires that a woman who can prove her husband is unfaithful must kill him that very day. No woman would dream of disobeying this law.
One day, the queen, who is known to be infallible, visits the village. She announces that at least one husband has been unfaithful.
What happens?
+ Show Spoiler +On the 50th day after the queen visits all wives kill their husbands. Same reasoning as the red eye blue eye monk question.
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A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
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Oh, and one my philosophy teacher gave me;
The following sentence is false. The preceding sentence is true. Are these sentences true or false?
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On December 06 2010 07:32 TymerA wrote: A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
Is it? + Show Spoiler +
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United States4053 Posts
On December 06 2010 07:32 TymerA wrote: A man was to be sentenced, and the judge told him, "You may make a statement. If it is true, I'll sentence you to four years in prison. If it is false, I'll sentence you to six years in prison." After the man made his statement, the judge decided to let him go free. What did the man say?
+ Show Spoiler + My sentence will be 6 years in prison.
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On December 06 2010 07:23 ixi.genocide wrote: The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated).
The awesome thing about the riddle is that even for veteran problemsolvers, it seems unsolvable, even though it is, without any stupd gimick.
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That one is kind of simple, but can has the potential to originate serious headaches.
Man of the 40th floor
+ Show Spoiler +A man lives on a 40th floor. To go down the building, he takes up the elevator and goes down straight to the B floor. When it goes up, he stops on the 35th floor, and walks up the stairs to the 40th floor. However, on rainy days he gets directly from B to 40th. Why?
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On December 06 2010 07:36 DumEN wrote:Show nested quote +On December 06 2010 07:23 ixi.genocide wrote: The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated). The awesome thing about the riddle is that even for veteran problemsolvers, it seems unsolvable, even though it is, without any stupd gimick.
Well then please give the answer before my head explodes!
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United States4053 Posts
On December 06 2010 07:37 Golondrin wrote:That one is kind of simple, but can has the potential to originate serious headaches. Man of the 40th floor+ Show Spoiler +A man lives on a 40th floor. To go down the building, he takes up the elevator and goes down straight to the B floor. When it goes up, he stops on the 35th floor, and walks up the stairs to the 40th floor. However, on rainy days he gets directly from B to 40th. Why? Total guess. + Show Spoiler + he's a midget, so he can't reach the 40 button, but on rainy days when he wears rain boots and has an umbrella, he's just tall enough to reach it
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On December 06 2010 07:41 infinitestory wrote:Show nested quote +On December 06 2010 07:37 Golondrin wrote:That one is kind of simple, but can has the potential to originate serious headaches. Man of the 40th floor+ Show Spoiler +A man lives on a 40th floor. To go down the building, he takes up the elevator and goes down straight to the B floor. When it goes up, he stops on the 35th floor, and walks up the stairs to the 40th floor. However, on rainy days he gets directly from B to 40th. Why? Total guess. + Show Spoiler + he's a midget, so he can't reach the 40 button, but on rainy days when he wears rain boots and has an umbrella, he's just tall enough to reach it
+ Show Spoiler +An umbrella can do the trick aswell So fast!
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On December 06 2010 07:32 Trion wrote:Show nested quote +On December 06 2010 06:49 xlat wrote:Love the thread, huge fan of this stuff. One of my favourites goes like this (didnt see it posted yet): + Show Spoiler +Every man in a village of fifty couples has been unfaithful to his wife. Every woman in the village instantly knows when a man other than her husband has philandered (you know how small towns are) but not when her own husband has ("always the last to know"). The village's notolerance adultery statute requires that a woman who can prove her husband is unfaithful must kill him that very day. No woman would dream of disobeying this law.
One day, the queen, who is known to be infallible, visits the village. She announces that at least one husband has been unfaithful.
What happens? + Show Spoiler +On the 50th day after the queen visits all wives kill their husbands. Some reasoning as the red eye blue eye monk question.
Yeah, correct. But for some reason I always have a much harder time both thinking about and putting into words the solution to the couples one compared to any of the other similar ones involving colors (there are many of them)
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On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
This doesn't seem too hard, so I probably got it wrong :D
+ Show Spoiler + Ask "Are you the middle daughter?" Youngest - Yes Middle - Yes/No Oldest - No
If you get 2x Yes, marry the "No" If you get 2x No, marry the "Yes"
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On December 06 2010 07:40 Trion wrote:Show nested quote +On December 06 2010 07:36 DumEN wrote:On December 06 2010 07:23 ixi.genocide wrote: The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated). The awesome thing about the riddle is that even for veteran problemsolvers, it seems unsolvable, even though it is, without any stupd gimick. Well then please give the answer before my head explodes!
Answer: + Show Spoiler +Really? You don't want to figure it out yourself? + Show Spoiler +Let's name the daughters A, B, and C. The question is: + Show Spoiler +[Asked to daughter A] "Is daugther B younger than daughter C?" To avoid choosing the middle daughter, always marry the one that is indicated to be younger.
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On December 06 2010 07:43 midgettoes wrote:This doesn't seem too hard, so I probably got it wrong :D + Show Spoiler + Ask "Are you the middle daughter?" Youngest - Yes Middle - Yes/No Oldest - No
If you get 2x Yes, marry the "No" If you get 2x No, marry the "Yes"
Haha , thats exactly what I had. But you are only allowed to ask >1< question to a random daughter, not 1 each.
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On December 06 2010 07:43 midgettoes wrote:Show nested quote +On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
This doesn't seem too hard, so I probably got it wrong :D + Show Spoiler + Ask "Are you the middle daughter?" Youngest - Yes Middle - Yes/No Oldest - No
If you get 2x Yes, marry the "No" If you get 2x No, marry the "Yes"
But you can only ask one question to one random daughter
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On December 06 2010 07:43 midgettoes wrote:Show nested quote +On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
This doesn't seem too hard, so I probably got it wrong :D + Show Spoiler + Ask "Are you the middle daughter?" Youngest - Yes Middle - Yes/No Oldest - No
If you get 2x Yes, marry the "No" If you get 2x No, marry the "Yes"
+ Show Spoiler +You can only ask it one time, to one daughter.
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ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats
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On December 06 2010 07:45 DumEN wrote:Show nested quote +On December 06 2010 07:40 Trion wrote:On December 06 2010 07:36 DumEN wrote:On December 06 2010 07:23 ixi.genocide wrote: The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated). The awesome thing about the riddle is that even for veteran problemsolvers, it seems unsolvable, even though it is, without any stupd gimick. Well then please give the answer before my head explodes! Answer: + Show Spoiler +Really? You don't want to figure it out yourself? + Show Spoiler +Let's name the daughters A, B, and C. The question is: + Show Spoiler +[Asked to daughter A] "Is daugther B younger than daughter C?" To avoid choosing the middle daughter, always marry the one that is indicated to be younger.
Thank you. Seems so easy now! Doh!
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On December 06 2010 07:47 DumEN wrote:Show nested quote +On December 06 2010 07:43 midgettoes wrote:On December 06 2010 06:22 DumEN wrote:I really like this one: Three Princesses+ Show Spoiler +You are a noble knight, who have just slain a vicious dragon. To reward you, the king will let you marry one of his three daughters. The daughters all look the same, and it is impossible to know them apart, unless you ask them. However, their personalities differ, a lot:
The youngest daughter always lie. The middle daughter lie half the time. The oldest daugters never lies.
After some concidering, you decide that it would be OK to live with either the youngest, or the eldest, because at least you know what they think.
When the time comes to choose, the king only allows you to ask one(1) "yes/no"-question to any one of the three daughters, and you don't know who is who.
What do you ask, to make sure you can make a choice, confident you will not choose the middle daughter?
This doesn't seem too hard, so I probably got it wrong :D + Show Spoiler + Ask "Are you the middle daughter?" Youngest - Yes Middle - Yes/No Oldest - No
If you get 2x Yes, marry the "No" If you get 2x No, marry the "Yes"
+ Show Spoiler +You can only ask it one time, to one daughter.
Looks like I shouldn't be answering things too early in the morning >.> woops can't read
+ Show Spoiler +In that case I need to think about it more... but maybe after some sleep
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On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats
+ Show Spoiler +If B and C were wearing white hats then A would know he had a black hat and would of answered with yes.
C has a black hat.
If C had a white hat and B was also wearing a white hat A would know he had a black hat. If C was wearing a white hat and B had a black hat B would know he had a black hat since he knew that A did not know what color hat he had. Since neither B or A knows what color hat they have C knows he must have a black hat.
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United States4053 Posts
On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats + Show Spoiler + A can see both B & C. A would obviously know his hat is black if B and C both had white hats. A doesn't know his hat color. B and C cannot both have white hats. B can see C. B has made the above deductions. B would know his hat is black if C had a white hat, because B and C do not both have white hats. B doesn't know his hat color. C must have a black hat. C has made the above deductions. C answers correctly that he has a black hat.
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On December 06 2010 07:56 infinitestory wrote:Show nested quote +On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats + Show Spoiler + A can see both B & C. A would obviously know his hat is black if B and C both had white hats. A doesn't know his hat color. B and C cannot both have white hats. B can see C. B has made the above deductions. B would know his hat is black if C had a white hat, because B and C do not both have white hats. B doesn't know his hat color. C must have a black hat. C has made the above deductions. C answers correctly that he has a black hat.
+ Show Spoiler +but B & C can both have black hats
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You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 3 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
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United States4053 Posts
On December 06 2010 07:58 Mobius wrote:Show nested quote +On December 06 2010 07:56 infinitestory wrote:On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats + Show Spoiler + A can see both B & C. A would obviously know his hat is black if B and C both had white hats. A doesn't know his hat color. B and C cannot both have white hats. B can see C. B has made the above deductions. B would know his hat is black if C had a white hat, because B and C do not both have white hats. B doesn't know his hat color. C must have a black hat. C has made the above deductions. C answers correctly that he has a black hat.
+ Show Spoiler +but B & C can both have black hats + Show Spoiler + in which case C's hat is still black
On December 06 2010 07:59 TymerA wrote: You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 4 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler +
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On December 06 2010 07:59 TymerA wrote:+ Show Spoiler [3 Doors] + You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 4 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler +Door 2, since there is a power outage.
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@ Tymera really, don't do something where you have to go to the beginning of it to figure it out.
damnit I was beaten by two people
Here's a good one that I got.
So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head.
When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why?
Answer + Show Spoiler +
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On December 06 2010 08:00 infinitestory wrote:Show nested quote +On December 06 2010 07:58 Mobius wrote:On December 06 2010 07:56 infinitestory wrote:On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats + Show Spoiler + A can see both B & C. A would obviously know his hat is black if B and C both had white hats. A doesn't know his hat color. B and C cannot both have white hats. B can see C. B has made the above deductions. B would know his hat is black if C had a white hat, because B and C do not both have white hats. B doesn't know his hat color. C must have a black hat. C has made the above deductions. C answers correctly that he has a black hat.
+ Show Spoiler +but B & C can both have black hats + Show Spoiler + in which case C's hat is still black
Show nested quote +On December 06 2010 07:59 TymerA wrote: You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 4 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler +
Typo.
The other guys got it.
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On December 06 2010 08:03 Thoreezhea wrote:@ Tymera really, don't do something where you have to go to the beginning of it to figure it out. damnit I was beaten by two people Here's a good one that I got. So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? Answer + Show Spoiler + I'm not sure if this is right, but it sounds reasonable + Show Spoiler + The noose around his neck chokes him and keeps the poison from entering his stomach, the fire burns through the rope, the makes puts him in to much pain or somehow inhibits him to correctly pulling the trigger. Since you did not specify where this was I can assume he is over a bridge that is over water, so he falls into the water and the fire goes out.
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United States4053 Posts
On December 06 2010 08:15 Trion wrote:Show nested quote +On December 06 2010 08:03 Thoreezhea wrote:@ Tymera really, don't do something where you have to go to the beginning of it to figure it out. damnit I was beaten by two people Here's a good one that I got. So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? Answer + Show Spoiler + I'm not sure if this is right, but it sounds reasonable + Show Spoiler + The noose around his neck chokes him and keeps the poison from entering his stomach, the fire burns through the rope, the makes puts him in to much pain or somehow inhibits him to correctly pulling the trigger. Since you did not specify where this was I can assume he is over a bridge that is over water, so he falls into the water and the fire goes out.
+ Show Spoiler + Since it says he does pull the trigger, I guess you should say the gun is empty
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What part of a woman's attire most stokes a man's desire?
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As I said the gun is loaded.
and we can also assume that he didn't swallow the poison instantly before jumping. Wtf?
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On December 06 2010 08:15 Trion wrote:Show nested quote +On December 06 2010 08:03 Thoreezhea wrote:@ Tymera really, don't do something where you have to go to the beginning of it to figure it out. damnit I was beaten by two people Here's a good one that I got. So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? Answer + Show Spoiler + I'm not sure if this is right, but it sounds reasonable + Show Spoiler + The noose around his neck chokes him and keeps the poison from entering his stomach, the fire burns through the rope, the makes puts him in to much pain or somehow inhibits him to correctly pulling the trigger. Since you did not specify where this was I can assume he is over a bridge that is over water, so he falls into the water and the fire goes out.
+ Show Spoiler +Thor messed up the order and the wording, but that's the answer.
The man put vial of poison in one hand, other hand a loaded gun, then stood on a stool and put on the noose. He drinks the poison, then kicks the spool and shoots. When he shoots where his head should be, him falling hits the noose, and he falls off and spits out the poison. Or some variation.
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On December 06 2010 08:00 infinitestory wrote:Show nested quote +On December 06 2010 07:58 Mobius wrote:On December 06 2010 07:56 infinitestory wrote:On December 06 2010 07:48 Mobius wrote:ive read the first 2 pages about the 3 hats + Show Spoiler +and im pretty sure they're all wrong.. If C could see A&B's hats, then yea he would know he is wearing a black hat.. But he doesnt know they're both wearing white hats + Show Spoiler + A can see both B & C. A would obviously know his hat is black if B and C both had white hats. A doesn't know his hat color. B and C cannot both have white hats. B can see C. B has made the above deductions. B would know his hat is black if C had a white hat, because B and C do not both have white hats. B doesn't know his hat color. C must have a black hat. C has made the above deductions. C answers correctly that he has a black hat.
+ Show Spoiler +but B & C can both have black hats + Show Spoiler + in which case C's hat is still black
Show nested quote +On December 06 2010 07:59 TymerA wrote: You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 4 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler + I GET IT! ahah i feel really stupid now :'(
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On December 06 2010 07:45 DumEN wrote:Show nested quote +On December 06 2010 07:40 Trion wrote:On December 06 2010 07:36 DumEN wrote:On December 06 2010 07:23 ixi.genocide wrote: The 3 princess question is unsolvable if there isn't a pattern to the middle daughters lying habits. If there was a pattern to it you could solve but there isn't (as far as the question has been stated). The awesome thing about the riddle is that even for veteran problemsolvers, it seems unsolvable, even though it is, without any stupd gimick. Well then please give the answer before my head explodes! Answer: + Show Spoiler +Really? You don't want to figure it out yourself? + Show Spoiler +Let's name the daughters A, B, and C. The question is: + Show Spoiler +[Asked to daughter A] "Is daugther B younger than daughter C?" To avoid choosing the middle daughter, always marry the one that is indicated to be younger. + Show Spoiler +since when do the daughters know who is who? it was stated that its impossible to tell them apart so you can't even ask this question =/
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On December 06 2010 08:21 Thoreezhea wrote: Wrong again blisse. The answer is different.
I will post the answer when we get to..... say 12 pages and no one has figured it out.
+ Show Spoiler +? Look again thanks. "Swallow seawater and vomits out the poison" doesn't count...
and where does the 'again' come from?
So let's see again... Swallow poison, bullet cuts rope, falls off cliff, into sea, swallows seawater, vomits out poison, puts out fire...
At below, read again...
+ Show Spoiler +The noose cut off circulation, fall off and impact makes you spit out what's in the mouth, I didn't say vomit... the doesn't count is because you don't mention water in the riddle, just a rock. Rock =\= water?
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On December 06 2010 08:22 Blisse wrote:Show nested quote +On December 06 2010 08:21 Thoreezhea wrote: Wrong again blisse. The answer is different.
I will post the answer when we get to..... say 12 pages and no one has figured it out. + Show Spoiler +? Look again thanks. "Swallow seawater and vomits out the poison" doesn't count...
and where does the 'again' come from?
again cuz some other person tried.
that doesn't count? huh. like he choked and vomited it up is more logical.
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you got it in your edit.
+ Show Spoiler + If I had mentioned water it would have been to easy, think out of the box right? sign of respect, I didn't realize there were so many different variations. jesus, anyway I just wanted to mention That the dude eventually did die of hypothermia in a french hospital. they couldn't save him. and yes, it did happen.
there was a tree on a cliff and a large boulder below it, jacque did everything that i said and then wen he jumped of the rock and jerked back throwing his aim off, and shattering the hook that was attached to the tree (as bullets don't cut rope), he fell of the cliff, the water put out the fire, In his panic at these unexpected events he swallowed a massive amount of sea water, which caused him to vomit up all of his stomach contents.
some more.
a penniless man made a beautiful metal sculpture out of scavenged materials. he soon died because of this.The materials used were perfectly safe and he had access to the basic tools to create it. Why?
a butterfly fell down and a man was seriously injured. The butterfly was not a valve or anything of innately harmful nature. Why?
A cop is at an intersection on a motorcycle during a red light and 4 teenagers in a car race past him through the intersection at 50 mph. the cop did not attempt to detain them or pursue them in any way. Why?
A famous man undresses for bed and millions lose their job, Why?
a married and maritally content woman sends 1000 anonymous valentine cards to local men. She didn't hand deliver them in any form and sent them through the postal service. why?
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United States4053 Posts
On December 06 2010 08:26 Thoreezhea wrote:Show nested quote +On December 06 2010 08:22 Blisse wrote:On December 06 2010 08:21 Thoreezhea wrote: Wrong again blisse. The answer is different.
I will post the answer when we get to..... say 12 pages and no one has figured it out. + Show Spoiler +? Look again thanks. "Swallow seawater and vomits out the poison" doesn't count...
and where does the 'again' come from? again cuz some other person tried. that doesn't count? huh. like he choked and vomited it up is more logical. + Show Spoiler + The noose choked the gun, preventing the bullet from exiting. This conveniently caused the gun to explode. Not only did the explosion produce a shock wave that put out the fire, the fumes caused the person to cough up the poison. Lastly, the poison happened to be mildly acidic, so after some of it landed on the noose, the noose also melted. He then went to get a drink of water from a nearby lake.
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On December 06 2010 08:40 infinitestory wrote:Show nested quote +On December 06 2010 08:26 Thoreezhea wrote:On December 06 2010 08:22 Blisse wrote:On December 06 2010 08:21 Thoreezhea wrote: Wrong again blisse. The answer is different.
I will post the answer when we get to..... say 12 pages and no one has figured it out. + Show Spoiler +? Look again thanks. "Swallow seawater and vomits out the poison" doesn't count...
and where does the 'again' come from? again cuz some other person tried. that doesn't count? huh. like he choked and vomited it up is more logical. + Show Spoiler + The noose choked the gun, preventing the bullet from exiting. This conveniently caused the gun to explode. Not only did the explosion produce a shock wave that put out the fire, the fumes caused the person to cough up the poison. Lastly, the poison happened to be mildly acidic, so after some of it landed on the noose, the noose also melted. He then went to get a drink of water from a nearby lake.
Holy shit dude. you crazy's are pretty creative.
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On December 06 2010 07:59 TymerA wrote: You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 4 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler +Isn't the answer supposed to be door 2 because the power went down? LOL. not door 4
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So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head.
When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why?
+ Show Spoiler + poison is not lethal? the noose reaches the ground so there is no tension on the rope? the gun is a water gun so it puts out the fire?
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On December 06 2010 08:54 aztrorisk wrote:So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? + Show Spoiler + poison is not lethal? the noose reaches the ground so there is no tension on the rope? the gun is a water gun so it puts out the fire?
he is DETERMINED to kill himself, he would make sure that doesn't happen.
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United States4053 Posts
On December 06 2010 08:39 Thoreezhea wrote:you got it in your edit. + Show Spoiler + If I had mentioned water it would have been to easy, think out of the box right? sign of respect, I didn't realize there were so many different variations. jesus, anyway I just wanted to mention That the dude eventually did die of hypothermia in a french hospital. they couldn't save him. and yes, it did happen.
there was a tree on a cliff and a large boulder below it, jacque did everything that i said and then wen he jumped of the rock and jerked back throwing his aim off, and shattering the hook that was attached to the tree (as bullets don't cut rope), he fell of the cliff, the water put out the fire, In his panic at these unexpected events he swallowed a massive amount of sea water, which caused him to vomit up all of his stomach contents. some more. a penniless man made a beautiful metal sculpture out of scavenged materials. he soon died because of this. Why? a butterfly fell down and a man was seriously injured. Why? A cop is at an intersection during a red light and 4 teenagers in a car race past him through the intersection at 50 mph. the cop did not attempt to detain them or pursue them in any way. Why? A man undresses for bed and millions lose their job, Why? a married and maritally content woman sends 1000 anonymous valentine cards to local men, why? 1. + Show Spoiler +the scavenged materials were coins he had found... then he ran out of food 2. + Show Spoiler +it was a butterfly sculpture, and it landed on the guy 3. + Show Spoiler + 4. + Show Spoiler + 5. + Show Spoiler +she's advertising a valentine card-giving service
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On December 06 2010 07:35 TymerA wrote: Oh, and one my philosophy teacher gave me;
The following sentence is false. The preceding sentence is true. Are these sentences true or false?
"No."
The third one is a question, and questions are not the sort of statements that are true or false, just like commands and exclamations. (Yes, I know that the intent was probably just to look at the first two and notice that you can't consistently assign both of them truth-values...)
Edit: Here's one of my favorites, might as well contribute to the thread. (Even if this is a pretty well-known one..) You wake up in a room which is empty except for two seemingly identical iron rods, and a note which says that exactly one of them is magnetized. You must determine which one it is, but have nothing but your hands and your wits to help you -- no metallic objects of any kind, no means of breaking or bending the bars, no measuring instruments of any kind, and no access to the outside world.
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On December 06 2010 08:59 Thoreezhea wrote:Show nested quote +On December 06 2010 08:54 aztrorisk wrote:So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? + Show Spoiler + poison is not lethal? the noose reaches the ground so there is no tension on the rope? the gun is a water gun so it puts out the fire?
he is DETERMINED to kill himself, he would make sure that doesn't happen.
+ Show Spoiler +Ok, my second guess. He jumps off the rock then fires the gun. The bullet hits the noose and cuts it off (because it was originally at his head but when he jumps, it is now above his head). As a result, he falls off the rock and he falls into a lake (which puts out the fire) or a wind blows so strongly that it puts out the fire. The motion forces him to throw up most of the poison
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United States4053 Posts
On December 06 2010 09:09 Iranon wrote:Show nested quote +On December 06 2010 07:35 TymerA wrote: Oh, and one my philosophy teacher gave me;
The following sentence is false. The preceding sentence is true. Are these sentences true or false?
"No." The third one is a question, and questions are not the sort of statements that are true or false, just like commands and exclamations. (Yes, I know that the intent was probably just to look at the first two and notice that you can't consistently assign both of them truth-values...) Edit: Here's one of my favorites, might as well contribute to the thread. (Even if this is a pretty well-known one..) You wake up in a room which is empty except for two seemingly identical iron rods, and a note which says that exactly one of them is magnetized. You must determine which one it is, but have nothing but your hands and your wits to help you -- no metallic objects of any kind, no means of breaking or bending the bars, no measuring instruments of any kind, and no access to the outside world. + Show Spoiler + Balance the rods one at a time horizontally on your finger or head, and see which one rotates.
That seems too simple, can't be right.
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On December 06 2010 08:39 Thoreezhea wrote:you got it in your edit. + Show Spoiler + If I had mentioned water it would have been to easy, think out of the box right? sign of respect, I didn't realize there were so many different variations. jesus, anyway I just wanted to mention That the dude eventually did die of hypothermia in a french hospital. they couldn't save him. and yes, it did happen.
there was a tree on a cliff and a large boulder below it, jacque did everything that i said and then wen he jumped of the rock and jerked back throwing his aim off, and shattering the hook that was attached to the tree (as bullets don't cut rope), he fell of the cliff, the water put out the fire, In his panic at these unexpected events he swallowed a massive amount of sea water, which caused him to vomit up all of his stomach contents. some more. a penniless man made a beautiful metal sculpture out of scavenged materials. he soon died because of this. Why? a butterfly fell down and a man was seriously injured. Why? A cop is at an intersection during a red light and 4 teenagers in a car race past him through the intersection at 50 mph. the cop did not attempt to detain them or pursue them in any way. Why? A man undresses for bed and millions lose their job, Why? a married and maritally content woman sends 1000 anonymous valentine cards to local men, why? + Show Spoiler +
1.) Being penniless, the man did not have access to tools and so during the sculpture's construction, cut himself and got blood poisoning via lead/rust/arsenic/whatever was in the scavenged metals.
2.) This butterfly here could refer to the butterfly boat or the butterfly valve, either of which could injure a man if it fell on him.
3.) The teenagers had a green light. There's always a red light at any intersection that has a light.
4.) The two are unrelated? The millions losing jobs is because of an economic downturn. Men normally undress for bed. They could easily happen at the same time.
5.) She's a postwoman.
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On December 06 2010 09:17 infinitestory wrote:Show nested quote +On December 06 2010 09:09 Iranon wrote:On December 06 2010 07:35 TymerA wrote: Oh, and one my philosophy teacher gave me;
The following sentence is false. The preceding sentence is true. Are these sentences true or false?
"No." The third one is a question, and questions are not the sort of statements that are true or false, just like commands and exclamations. (Yes, I know that the intent was probably just to look at the first two and notice that you can't consistently assign both of them truth-values...) Edit: Here's one of my favorites, might as well contribute to the thread. (Even if this is a pretty well-known one..) You wake up in a room which is empty except for two seemingly identical iron rods, and a note which says that exactly one of them is magnetized. You must determine which one it is, but have nothing but your hands and your wits to help you -- no metallic objects of any kind, no means of breaking or bending the bars, no measuring instruments of any kind, and no access to the outside world. + Show Spoiler + Balance the rods one at a time horizontally on your finger or head, and see which one rotates.
That seems too simple, can't be right. haha, I like yours. + Show Spoiler +make a "T" shape with the two bars. If they atract, the shaft is the magnet. If they don't the head is the magnet.
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On December 06 2010 09:13 aztrorisk wrote:Show nested quote +On December 06 2010 08:59 Thoreezhea wrote:On December 06 2010 08:54 aztrorisk wrote:So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? + Show Spoiler + poison is not lethal? the noose reaches the ground so there is no tension on the rope? the gun is a water gun so it puts out the fire?
he is DETERMINED to kill himself, he would make sure that doesn't happen. + Show Spoiler +Ok, my second guess. He jumps off the rock then fires the gun. The bullet hits the noose and cuts it off (because it was originally at his head but when he jumps, it is now above his head). As a result, he falls off the rock and he falls into a lake (which puts out the fire) or a wind blows so strongly that it puts out the fire. The motion forces him to throw up most of the poison
+ Show Spoiler + correct, but it is seawater that he swallowed and not motion sickness that causes him to vomit. Good job.
Phael was correct on three
all of the others are completely accurate in that they could EASILY happen. I thought that the book I had was very unclear and stupid so I will edit my riddles. However, I would like to see someone guess the answers that the book thinks are plausible.
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It happens once in a minute, twice in a week, and once in a year. What is it?
How can half of 12 be 7?
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It happens once in a minute, twice in a week, and once in a year. What is it? + Show Spoiler +
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United States4053 Posts
On December 06 2010 10:13 Trion wrote: It happens once in a minute, twice in a week, and once in a year. What is it?
How can half of 12 be 7? + Show Spoiler +
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A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water.
Is there more wine in the water or is there more water in the wine?
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On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of win.
I loled
I am confused.
jug A is the water, Jug B is the wine, was the jug poured into the jug, or was it mixed in a glass or what? Was everything in a glass?
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On December 06 2010 10:33 Thoreezhea wrote:Show nested quote +On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of win. I loled
Oops.
Fixed.
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On December 06 2010 10:33 Thoreezhea wrote:Show nested quote +On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of win. I loled I am confused. jug A is the water, Jug B is the wine, was the jug poured into the jug, or was it mixed in a glass or what?
A glass of water was taken out of the first jug and poured into the second, then a up from the second was poured into the first. It makes sense, just try rereading.
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On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water.
Is there more wine in the water or is there more water in the wine?
+ Show Spoiler + As long as the two jugs hold the same volume of liquid, then the amount of wine in the water jugmust be equal to the amount of water in the wine jug.
Logically, if there is X amount of wine in the water jug, then X amount of water must've been displaced into the wine jug.
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+ Show Spoiler + There is more water in the wine.
Treat the problem like this. say that the jug has 4 letters of fluid. so currently there is (A = water and B = wine)
AAAA = BBBB
when you take 1 glass of water say 2 letters and pour it into the wine you have
AA and BBBBAA
When you take some A/B mixture from that and put it in the water it is changed again.
Keep in mind that there are 1 parts water to 2 parts wine
AAB 1/2 A and BBBA 1/2 A There is more water in the wine. Edit, I'm stupid.
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On December 06 2010 10:54 Thoreezhea wrote:+ Show Spoiler + There is more water in the wine.
Treat the problem like this. say that the jug has 4 letters of fluid. so currently there is (A = water and B = wine)
AAAA = BBBB
when you take 1 glass of water say 2 letters and pour it into the wine you have
AA and BBBBAA
When you take some A/B mixture from that and put it in the water it is changed again.
Keep in mind that there are 1 parts water to 2 parts wine
AAB 1/2 A and BBBA 1/2 A There is more water in the wine.
Um ...
+ Show Spoiler + You say a glass of water = 2 letters? Then on the second pouring, you only scooped up 1.5 letters.
The mixture would be 2:1 B to A, so you'd be scooping up 4/3 letters of B and 2/3 letters of A, resulting in:
2 2/3A + 1 1/3B vs 2 2/3 B + 1 1/3 A.
Numerically, if you had 60 cc of water and 60 cc of wine, first scoop:
30 cc of water vs 60 wine + 30 water second scoop would contain 20 cc wine + 10 cc water resulting in:
40cc water, 20cc wine vs 40cc wine, 20cc water
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+ Show Spoiler [water and wine jugs] +The quantities of water in wine and wine in water are equal. This is due to the fact that there is half a glass of water and half a glass of wine (mixed) replaced for a full glass of water in the water jug.
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A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
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On December 06 2010 11:13 Trion wrote: A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
+ Show Spoiler +
Left bank: Fox, Corn | Boat: farmer, goose | right bank: 0 Left bank: Fox, Corn | Boat: farmer | right bank: goose Left bank: Fox | Boat: farmer, corn | right bank: goose Left bank: Fox | Boat: farmer, goose | right bank: corn Left bank: Goose | Boat: farmer, fox | right bank: corn Left bank: Goose | Boat: farmer | right bank: fox, corn Left bank: 0 | Boat: farmer, goose | right bank: fox, corn Left bank: 0 | Boat: 0 | right bank: fox, goose, corn
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Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural.
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. Meh TL page is too narrow to write my proof.
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On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler + The first part is simple, as there's no possibility of making it under 7 turns, but there are 7 +6+5+4+3+2+1= 28 paths to do it, depending on the order of your flips. Do consider that if the coins starts as 4 heads/tails, it'll still count as if you've done it. Hence the 7 turns. (To make it more clear, if the coins starts as head, tails, head, tails, and you choose to flip the tails in this setup(whilst not knowing they were tails): the task would be done in the first turn. However, if they started out as head head tails head; it'd set them to head tails tails tails, in which you flip the head for that setup and clear the task in the second turn. Doing this in order so every possibility is given would take 7 turns whatever path you choose to take, as no path would make 2 or more setups to get cleared in one turn.
The thing about the second part is that when he turns the table, the possibilities of your path might or might not change. So out of these 28 paths you have to find the one that gets it right every turn no matter what order the coins are in at the time. Which to me sounds impossible. Therefore I figure it's a trick question, the blind man can tell which coins are heads or tails with his fingertips, so he only needs one turn.
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platypuses0
United States44 Posts
On December 06 2010 10:13 Trion wrote: How can half of 12 be 7? + Show Spoiler +12 in Roman numerals is XII, if you cut that in half horizontally you get VII which is 7.
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If I recall correctly there is a solution to the blind man even with a rotating board, but I'm not 100% sure it's the exact same moves. You simply need to flip certain coins such that the orientation won't matter for your next step, as in you continuously see different scenarios.
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edit reposted : please remove
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0
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On December 06 2010 13:53 ieatpasta wrote:Show nested quote +On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0
It says prove there ISNT a solution. You proved that there is one.
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Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it?
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United States4053 Posts
On December 06 2010 14:17 ieatpasta wrote: Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it? + Show Spoiler +
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On December 06 2010 14:17 ieatpasta wrote: Riddle: Poor people have it. Rich people need it. It is greater than God and more evil than the Devil. If you eat it you will die. What is it?
+ Show Spoiler + The answer is "nothing"
poor poeple have nothing rich poeple need nothing . nothing is greater than god . nothing is more evil than the devil . if you eat nothing you will die
Edit: needed to add spoiler tag
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How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy)
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On December 06 2010 14:40 MrProphylactic wrote: How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy) + Show Spoiler +XIX - I = XX Roman Numerals
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On December 05 2010 13:30 Dr. ROCKZO wrote:Here's one of my favourites. Logic: + Show Spoiler +
+ Show Spoiler +I already know the solution, but I was thinking about it, and...
...pragmatically, you have a 100% chance of NOT DYING (death being the number one "motivator" as far as riddles are concerned). You weigh a given bag, and if it's normal weight, eat the candy--if not, eat from any of the other bags haha.
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On December 06 2010 14:05 zJayy962 wrote:Show nested quote +On December 06 2010 13:53 ieatpasta wrote:On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0 It says prove there ISNT a solution. You proved that there is one.
Not a solution. 0 isn't a natural number.
+ Show Spoiler + For those who aren't aware, this is known as Fermat's last theorem. He scribbled it in the margins of a book that he had thought of a marvelous and elegant solution to it, but the margin was too small to fit it in (hence an earlier joke "solution").
Mathematicians tried to prove it for hundreds of years, and it was only through reams of paper and years of computation that someone did it a few years back.
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On December 06 2010 11:13 Trion wrote: A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
Well the same Warlock that transformed his sheep into the corn and his cabbage into a goose. Could magic him across the river.
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On December 06 2010 08:03 Thoreezhea wrote:@ Tymera really, don't do something where you have to go to the beginning of it to figure it out. damnit I was beaten by two people Here's a good one that I got. So a man named Jacque is very determined to kill himself, he swallows poison, ties a noose around his neck and that to a tree, he sets himself on fire, then prepares to pull the trigger of a loaded gun he is holding up to his head. When he finally does jump of the rock he is standing on and pulls the trigger, He does not die. Why? Answer + Show Spoiler +
Unlikely, but total guess xD + Show Spoiler + The bullet is fired as he jumps, so the gun flails in the air and the bullet hits the rope, cutting the cord. He falls into the water? (It says he is on a rock... a large rock = cliff? with an overhanging tree perhaps...) which extinguishes the flames and he vomits out the poison after having drank some seawater. Seems like the only answer to me unless I'm missing something (possibly to do with the name... is it an alternate scenario that I'm not thinking about?)
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how many places are there on the earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere. also, the rotation of the earth has nothing to do with the solution; you can assume you're walking on a static sphere if that makes the problem less complicated to you.
Hint: + Show Spoiler +think you've figured it out? do you know that there's more than one? in fact, there are more than two. also note that walking north from the north pole (or south from the south pole) is illogical and therefore does not enter into the problem. all normal assumptions about directions will be used. Hint 2: + Show Spoiler +
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On December 06 2010 15:23 Phael wrote:Show nested quote +On December 06 2010 14:05 zJayy962 wrote:On December 06 2010 13:53 ieatpasta wrote:On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0 It says prove there ISNT a solution. You proved that there is one. Not a solution. 0 isn't a natural number. For those who aren't aware, this is known as Fermat's last theorem. He scribbled it in the margins of a book that he had thought of a marvelous and elegant solution to it, but the margin was too small to fit it in (hence an earlier joke "solution").
Mathematicians tried to prove it for hundreds of years, and it was only through reams of paper and years of computation that someone did it a few years back. Please don't say all that and not credit the man who found the solution! That guy's name is Andrew Wiles, and he just happens to chair the Math department at my university =)
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On December 06 2010 17:06 ixi.genocide wrote:how many places are there on the earth that one could walk one mile south, then one mile west, then one mile north and end up in the same spot? to be precise, let's assume the earth is a solid smooth sphere, so oceans and mountains and other such things do not exist. you can start at any point on the sphere. also, the rotation of the earth has nothing to do with the solution; you can assume you're walking on a static sphere if that makes the problem less complicated to you. Hint: + Show Spoiler +think you've figured it out? do you know that there's more than one? in fact, there are more than two. also note that walking north from the north pole (or south from the south pole) is illogical and therefore does not enter into the problem. all normal assumptions about directions will be used. Hint 2: + Show Spoiler + + Show Spoiler +1. north pole 2. infinently many points in the south pole; just find a circle that crcunrence that goes into one evenly, and go nortth one mile. The circle that is one mile north of the south pole is a start, but also circles are one mile a way from circles that have circunfruce of 1 mile, .5 mile, 1/3 mile, .25 mile... I hope I made that clear. I think the real riddle is what columbus has to do with this . I feel that I should contribute, so here a fraction one, yay fractions! father's will + Show Spoiler +In a father's will he says that he will leave 1/2 of his fortune to hiis eldest son, 1/3 to the middle son, and 1/9 to the youngest son. They split up land and wealth fairly easily, but a problem arises when they try to split up the father's prized stable, that contains 17 horses. 17 does not split evenly among 1/2, 1/3 and 1/9. One son surgests that the remaing horces be cut up, so that each son gets exactly what the father had willed, one son surgests they sell the remainder horses, spliting the money, and the final son surgests that some sort of trade is aranged so that one son will get more horses and less land or something like that. As the sons argue, a wise man rides up and over hears the argument and rides up to the men. With one action, he is able to find a solution that all of the sons are happy with. What did he do?
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This one kept me up for days
3 guys walk into a hotel and put down $10 each totalling $30 to get a room for the night. The manager hands over the key and the guys take the elevator up. Minutes later, the manager realized that the 3 guys overpaid - the actual price should have been $25. Therefore, he gives $5 to a boy asks him to take the $5 back up to the 3 guys. On the elevator, the boy wonders how 3 people can split $5, so he takes $2 and puts it in his own pocket and distrbutes the remaining $3 one dollar per person.
But wait, if each person just received a dollar back, it would technically mean that they each paid $9 instead of $10. 3 x 9 = 27 plus the $2 the boy stole = $29. What happened to the missing dollar?
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On December 06 2010 17:35 sksyen wrote: This one kept me up for nights
3 guys walk into a hotel and put down $10 each totalling $30 to get a room for the night. The manager hands over the key and the guys take the elevator up. Minutes later, the manager realized that the 3 guys overpaid - the actual price should have been $25. Therefore, he gives $5 to a boy asks him to take the $5 back up to the 3 guys. On the elevator, the boy wonders how 3 people can split $5, so he takes $2 and puts it in his own pocket and distrbutes the remaining $3 one dollar per person.
But wait, if each person just received a dollar back, it would technically mean that they each paid $9 instead of $10. 3 x 9 = 27 plus the $2 the boy stole = $29. What happened to the missing dollar?
this was in the pic thread within the past 2 weeks. + Show Spoiler +the two dolars that the boy stole are part of the 27 paid, so those two are being double counted. If you just want to look at where the original 30 dollars go, man 1 has 1, man 2 has 1, man 3 has 1, theif has 2, cash regester has 25. 1+1+1+2+25=30
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United States4053 Posts
On December 06 2010 17:35 sksyen wrote: This one kept me up for days
3 guys walk into a hotel and put down $10 each totalling $30 to get a room for the night. The manager hands over the key and the guys take the elevator up. Minutes later, the manager realized that the 3 guys overpaid - the actual price should have been $25. Therefore, he gives $5 to a boy asks him to take the $5 back up to the 3 guys. On the elevator, the boy wonders how 3 people can split $5, so he takes $2 and puts it in his own pocket and distrbutes the remaining $3 one dollar per person.
But wait, if each person just received a dollar back, it would technically mean that they each paid $9 instead of $10. 3 x 9 = 27 plus the $2 the boy stole = $29. What happened to the missing dollar? This one kept me up for days too. + Show Spoiler + The amount of money the 3 guys lost does indeed total to $27. However, the amount the manager has gained is $25. The extra $2 doesn't get added to $27, it's subtracted.
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On December 06 2010 17:35 sksyen wrote: This one kept me up for days
3 guys walk into a hotel and put down $10 each totalling $30 to get a room for the night. The manager hands over the key and the guys take the elevator up. Minutes later, the manager realized that the 3 guys overpaid - the actual price should have been $25. Therefore, he gives $5 to a boy asks him to take the $5 back up to the 3 guys. On the elevator, the boy wonders how 3 people can split $5, so he takes $2 and puts it in his own pocket and distrbutes the remaining $3 one dollar per person.
But wait, if each person just received a dollar back, it would technically mean that they each paid $9 instead of $10. 3 x 9 = 27 plus the $2 the boy stole = $29. What happened to the missing dollar? + Show Spoiler + umm, so they pay 30. Then, manager sends 5 back. The guy that the manager sends the money with takes 2 out of 5 and gives the remaining 3 to the boys. What's confusing about it? :o
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On December 06 2010 07:59 TymerA wrote: You've been arrested for murder. However, the goverment decided to send you to a sadistic tv game-show. During the show the power suddenly goes down. You try to escape, but accidently end up in the game-show's arena and you are standing in front of 3 doors.
Going trough door 1, you will end up in a open room with Mustard gas. Your head, arms and legs are not covered by clothing.
Going trough door 2 you will have to go trough water with a electric current going trough it because of a electric cable lying in the water at the far end of the room.
Going trough door 3 you will have to fight a group of very agressive bears. You can't fool them.
At the end of each room there is a door. If you make it to the door, your life will be spared.
What door should you go trough?
+ Show Spoiler +there is no power, go through door #2
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On December 06 2010 17:11 Wings wrote:Show nested quote +On December 06 2010 15:23 Phael wrote:On December 06 2010 14:05 zJayy962 wrote:On December 06 2010 13:53 ieatpasta wrote:On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0 It says prove there ISNT a solution. You proved that there is one. Not a solution. 0 isn't a natural number. For those who aren't aware, this is known as Fermat's last theorem. He scribbled it in the margins of a book that he had thought of a marvelous and elegant solution to it, but the margin was too small to fit it in (hence an earlier joke "solution").
Mathematicians tried to prove it for hundreds of years, and it was only through reams of paper and years of computation that someone did it a few years back. Please don't say all that and not credit the man who found the solution! That guy's name is Andrew Wiles, and he just happens to chair the Math department at my university =)
His solution had an error and was thus proven wrong, or do I remember the wrong guy?
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On December 06 2010 17:49 Ghazwan wrote:Show nested quote +On December 06 2010 17:35 sksyen wrote: This one kept me up for days
3 guys walk into a hotel and put down $10 each totalling $30 to get a room for the night. The manager hands over the key and the guys take the elevator up. Minutes later, the manager realized that the 3 guys overpaid - the actual price should have been $25. Therefore, he gives $5 to a boy asks him to take the $5 back up to the 3 guys. On the elevator, the boy wonders how 3 people can split $5, so he takes $2 and puts it in his own pocket and distrbutes the remaining $3 one dollar per person.
But wait, if each person just received a dollar back, it would technically mean that they each paid $9 instead of $10. 3 x 9 = 27 plus the $2 the boy stole = $29. What happened to the missing dollar? + Show Spoiler + umm, so they pay 30. Then, manager sends 5 back. The guy that the manager sends the money with takes 2 out of 5 and gives the remaining 3 to the boys. What's confusing about it? :o
The answer is false lol
the missing dollar didn't go anywhere because it does not exist you are not adding the 2 dollars to both sides of the equation properly, you are adding it to one side twice . The easiest way to visualize this is two sides of an algebraic equation . one side we will call the register side , the other the customer side . it all adds up, there is no extra dollar . 25 (at the register side) 25 in register plus 2 in the bellboys pocket = 27 plus 3 dollar refund = 30 . now if we add it up from the other side(customer side ) . 10 x 3 = 30 - 3 (refund ) =27 . 9x3=27 plus 3 = 30 . the only answer is : 25+2+2=27 is false
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Here is one. I hope my translation of the question will be easy to understand.:
A guy that works for the census bureau visits a house. The father of the house opens the door.
Census guy: How many children do you have? Father: I have three daughters. Census guy: Could you please tell me the exact ages of your daughters? Father: The product of their ages is 72 and the sum of their ages is our door number. Census guy: But, this is not enough information! Father: My oldest daughter loves horses. Census guy: Thank you.
How did he find the ages of the daughters and what's the house number?
Tip: + Show Spoiler +The guy sees the house number, but we do not know it.
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While I am at it, here is another one:
A courier is carrying a sack full of marbles. We know there were 300 marbles of five different colors in the sack. The courier though, while passing a river, slips the sack off of his hand and lets some marbles fall into the river. We know the amount of marbles that fell are at least 100. He tells us that out of the remaining marbles in the sack, 1/3rd of them are white, 1/4th of them are red, 1/5th of them are blue, 1/7th of them are yellow, and 1/9th of them are green. However, we also know that the courier is not so good with math and we are sure that one of the information he has given is wrong.
How many marbles of each color remained in the sack and how many marbles fell into the river?
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census guy + Show Spoiler +factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. so we have 1 1 72 = 74 1 2 36 = 39 1 3 24 = 28 1 4 18 = 23 1 6 12 = 19 1 8 9 = 18 2 2 18 = 22 (oh god im going to kill myself) 2 3 12 = 17 2 4 9 = 15 2 6 6 = 14 3 3 8 = 14 3 4 6 = 13 all the sums except 266 and 338 are unique, so it had to be 266 or 338 if the guy didnt know "oldest" means it's not 266, so the census guy knows its 338 actually did this myself, so forgive me if i missed anything
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On December 06 2010 18:42 Ghazwan wrote: While I am at it, here is another one:
A courier is carrying a sack full of marbles. We know there were 300 marbles of five different colors in the sack. The courier though, while passing a river, slips the sack off of his hand and lets some marbles fall into the river. We know the amount of marbles that fell are at least 100. He tells us that out of the remaining marbles in the sack, 1/3rd of them are white, 1/4th of them are red, 1/5th of them are blue, 1/7th of them are yellow, and 1/9th of them are green. However, we also know that the courier is not so good with math and we are sure that one of the information he has given is wrong.
How many marbles of each color remained in the sack and how many marbles fell into the river?
+ Show Spoiler + As far as I can tell, there are two solutions:
180 total marbles, 120 fell in the river
60 white 45 red 36 blue 19 yellow (error) 20 green
or 252 total marbles, 48 fell in the river 84 white 63 red 41 blue (error) 36 yellow 28 green
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On December 06 2010 18:56 5ahj4g wrote:census guy + Show Spoiler +factors of 72 are 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, and 72. so we have 1 1 72 = 74 1 2 36 = 39 1 3 24 = 28 1 4 18 = 23 1 6 12 = 19 1 8 9 = 18 2 2 18 = 22 (oh god im going to kill myself) 2 3 12 = 17 2 4 9 = 15 2 6 6 = 14 3 3 8 = 14 3 4 6 = 13 all the sums except 266 and 338 are unique, so it had to be 266 or 338 if the guy didnt know "oldest" means it's not 266, so the census guy knows its 338 actually did this myself, so forgive me if i missed anything
Well done
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On December 06 2010 18:56 Phael wrote:Show nested quote +On December 06 2010 18:42 Ghazwan wrote: While I am at it, here is another one:
A courier is carrying a sack full of marbles. We know there were 300 marbles of five different colors in the sack. The courier though, while passing a river, slips the sack off of his hand and lets some marbles fall into the river. We know the amount of marbles that fell are at least 100. He tells us that out of the remaining marbles in the sack, 1/3rd of them are white, 1/4th of them are red, 1/5th of them are blue, 1/7th of them are yellow, and 1/9th of them are green. However, we also know that the courier is not so good with math and we are sure that one of the information he has given is wrong.
How many marbles of each color remained in the sack and how many marbles fell into the river? + Show Spoiler + As far as I can tell, there are two solutions:
180 total marbles, 120 fell in the river
60 white 45 red 36 blue 19 yellow (error) 20 green
or 252 total marbles, 48 fell in the river 84 white 63 red 41 blue (error) 36 yellow 28 green
The first one is true, I am sorry I edited my question a bit late and added : "We know the amount of marbles that fell are at least 100."
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On December 06 2010 18:05 Ghazwan wrote:Show nested quote +On December 06 2010 17:11 Wings wrote:On December 06 2010 15:23 Phael wrote:On December 06 2010 14:05 zJayy962 wrote:On December 06 2010 13:53 ieatpasta wrote:On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural. + Show Spoiler +x=0 y=0 z=0 n>2 0^n + 0^n = 0^n 0+0=0 It says prove there ISNT a solution. You proved that there is one. Not a solution. 0 isn't a natural number. For those who aren't aware, this is known as Fermat's last theorem. He scribbled it in the margins of a book that he had thought of a marvelous and elegant solution to it, but the margin was too small to fit it in (hence an earlier joke "solution").
Mathematicians tried to prove it for hundreds of years, and it was only through reams of paper and years of computation that someone did it a few years back. Please don't say all that and not credit the man who found the solution! That guy's name is Andrew Wiles, and he just happens to chair the Math department at my university =) His solution had an error and was thus proven wrong, or do I remember the wrong guy?
His first solution didn't account for all possibilities. His second one did
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father's will + Show Spoiler +In a father's will he says that he will leave 1/2 of his fortune to hiis eldest son, 1/3 to the middle son, and 1/9 to the youngest son. They split up land and wealth fairly easily, but a problem arises when they try to split up the father's prized stable, that contains 17 horses. 17 does not split evenly among 1/2, 1/3 and 1/9. One son surgests that the remaing horces be cut up, so that each son gets exactly what the father had willed, one son surgests they sell the remainder horses, spliting the money, and the final son surgests that some sort of trade is aranged so that one son will get more horses and less land or something like that. As the sons argue, a wise man rides up and over hears the argument and rides up to the men. With one action, he is able to find a solution that all of the sons are happy with. What did he do? [/QUOTE]
I shall deem thy riddle with an answer, despite the typos =P + Show Spoiler + He gives them his horse. 17 does not divide by 9,3 and 2 but 18 sure does. After this, there should be ONE HORSE LEFT OVER since the question does not state what happens to the remaining 1/18th of the estate left over (1/2 + 1/3 +1/9 = 17/18). The wise man then takes that horse back and each son is happy with what he has received.
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barber paradox:
+ Show Spoiler +All the men in the village go to the barber to get their beard shaved and it is taboo to shave your own beard. But the barber is the only barber in the whole village. Who shaves his beard?
+ Show Spoiler +Just so you know when I heard it I was told there really isn't an answer
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On December 07 2010 00:09 SockMonkey wrote:barber paradox:+ Show Spoiler +All the men in the village go to the barber to get their beard shaved and it is taboo to shave your own beard. But the barber is the only barber in the whole village. Who shaves his beard?
+ Show Spoiler +Just so you know when I heard it I was told there really isn't an answer
the barber is a child doesnt grow facial hair or a woman or he grows a beard its really a horrible riddle.=
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On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. The problem is posted wrongly. You said the fire moves from A to B and the wind moves from A to B. In which case there is no solution.
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On December 06 2010 21:50 GQz wrote:
yea
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On December 06 2010 10:13 Trion wrote: It happens once in a minute, twice in a week, and once in a year. What is it?
How can half of 12 be 7?
+ Show Spoiler +the letter "E" 12 = XII ; 7 = VII in roman letters, cut through middle and voila
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On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water.
Is there more wine in the water or is there more water in the wine?
well, there is more water in wine than wine in water, assumed you wanted to say the glasses poured are the same quantity. + Show Spoiler +the 1st time you take just water, pour into wine the 2nd time you take wine+water and pour back into water
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On December 06 2010 11:13 Trion wrote: A farmer wants to cross a river with a fox, a sheep, and a cabbage. He must cross the river with a tiny boat. He could only take himself and one other - the fox, the goose, or the corn - at a time. He could not leave the fox alone with the goose or the goose alone with the corn. How does he get all safely over the stream?
foxes don't eat sheep, the aswer for the wolf version is simple
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On December 07 2010 01:56 jcroisdale wrote:Show nested quote +On December 07 2010 00:09 SockMonkey wrote:barber paradox:+ Show Spoiler +All the men in the village go to the barber to get their beard shaved and it is taboo to shave your own beard. But the barber is the only barber in the whole village. Who shaves his beard?
+ Show Spoiler +Just so you know when I heard it I was told there really isn't an answer the barber is a child doesnt grow facial hair or a woman or he grows a beard its really a horrible riddle.= Barber is a woman
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Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B.
Uh what? Regarding the answer...
+ Show Spoiler +How the hell will the wind blow the second fire towards the original fire, if it's blowing from A to B, while the fire is moving from A to B?
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On December 06 2010 12:02 EmptyW wrote:Show nested quote +On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler + The first part is simple, as there's no possibility of making it under 7 turns, but there are 7 +6+5+4+3+2+1= 28 paths to do it, depending on the order of your flips. Do consider that if the coins starts as 4 heads/tails, it'll still count as if you've done it. Hence the 7 turns. (To make it more clear, if the coins starts as head, tails, head, tails, and you choose to flip the tails in this setup(whilst not knowing they were tails): the task would be done in the first turn. However, if they started out as head head tails head; it'd set them to head tails tails tails, in which you flip the head for that setup and clear the task in the second turn. Doing this in order so every possibility is given would take 7 turns whatever path you choose to take, as no path would make 2 or more setups to get cleared in one turn.
The thing about the second part is that when he turns the table, the possibilities of your path might or might not change. So out of these 28 paths you have to find the one that gets it right every turn no matter what order the coins are in at the time. Which to me sounds impossible. Therefore I figure it's a trick question, the blind man can tell which coins are heads or tails with his fingertips, so he only needs one turn.
this is possible and it goes like this
logical data + Show Spoiler +A 2x2 checkers board can take 4 possible states 1: 4 coins same side - called from now on 4C 2: 1 coins same side (doesn't matter which 3 coins) - called from now on 1C 3: 2 adjacent coins same side - called from now on 2A 4: 2 diagonal coins same side - called from now on 2D -the 3 coins same side is the same as 3 coins same side There are couple of moves we are going to analyze: flip a single random coin = F1 flip two adjacent coins = F2A flip two diagonal coins = F2D No matter which position the board is in, F1(1C) = 4C or 2D or 2A F1(2A or 2D) = 1C F2A(1C) = 1C F2A(2A) = 4C or 2D F2A(2D) = 2D F2D(1C) = 1C F2D(2A) = 2A F2D(2D) = 4C Looking at the things above, we see that if we perform the moves F2D, F2A, F2D on a board with 2A or 2D we will eventually get 4C at some point The problem is that the board might be 1C then the solution is simple + Show Spoiler + F2D, F2A, F2D then F1 then F2D, F2A, F2D again - 7 moves in one of these, we are bound to encounter 4C
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On December 06 2010 17:31 hacklebeast wrote:father's will+ Show Spoiler +In a father's will he says that he will leave 1/2 of his fortune to hiis eldest son, 1/3 to the middle son, and 1/9 to the youngest son. They split up land and wealth fairly easily, but a problem arises when they try to split up the father's prized stable, that contains 17 horses. 17 does not split evenly among 1/2, 1/3 and 1/9. One son surgests that the remaing horces be cut up, so that each son gets exactly what the father had willed, one son surgests they sell the remainder horses, spliting the money, and the final son surgests that some sort of trade is aranged so that one son will get more horses and less land or something like that. As the sons argue, a wise man rides up and over hears the argument and rides up to the men. With one action, he is able to find a solution that all of the sons are happy with. What did he do? + Show Spoiler + The wise man had a horse. He said : "Add this horse to your late father's stable." (17 + 1 = 18) Eldest son took 1/2 from 18 = 9 Eldest son took 1/3 from 18 = 6 Eldest son took 1/9 from 18 = 2 These are 17 horses, the 18th horse belonged to the old man.
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Guys I still don't understand the Monty Hall paradox or whatever it's called, and I never have understood it.
If the Host knows that door 3 is a goat and shows it to you, how does it suddenly become a 66% chance of winning if you switch to door 2?
Wouldn't it just become a 50/50 chance to win? Ugh my brain's gonna explode from trying to figure this out. The 100 door example didn't help either, but I sort of understood it when one of the posters said that it would only make sense to stay if you had chosen correctly in the first place. Out of 100 doors if you knew you had a 1% chance to get the right door, and he opened 98 other doors, wouldn't your chance of being correct just increase until it was 50%?
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On December 05 2010 13:12 t3tsubo wrote:3 Hats [tag: logic]+ Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be?
+ Show Spoiler +His hat is Black. The guy in the back would only say "Yes" if the two in front of him were wearing either two black hats or one black hat and one white hat. Now, man B would only say "Yes" if he saw man A wearing a white hat. This is because if he is wearing a white hat, then the only possible hat that he could have on would be black. So, by hearing his answer of "no", the man in the front can be absolutely certain that he is wearing a black hat. This means that both men, B and C, are wearing black hats.
Definitely read this thinking that the unbolded ones were unanswered :S
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On December 07 2010 04:43 Pandonetho wrote: Guys I still don't understand the Monty Hall paradox or whatever it's called, and I never have understood it.
If the Host knows that door 3 is a goat and shows it to you, how does it suddenly become a 66% chance of winning if you switch to door 2?
Wouldn't it just become a 50/50 chance to win? Ugh my brain's gonna explode from trying to figure this out. The 100 door example didn't help either, but I sort of understood it when one of the posters said that it would only make sense to stay if you had chosen correctly in the first place. Out of 100 doors if you knew you had a 1% chance to get the right door, and he opened 98 other doors, wouldn't your chance of being correct just increase until it was 50%? + Show Spoiler +If you go in thinking "I will switch" the ONLY way you can lose is if you pick the door with the prize in the beginning. You have a 1/3 chance of picking that door, so a 1/3 chance of losing.
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On December 07 2010 04:03 MindRush wrote:Show nested quote +On December 06 2010 10:24 Trion wrote: A jug holds a quantity of water. Another jug holds an equal quantity of wine. A glass of water is taken from the first jug, poured into the wine, and the contents stirred. A glass of the mixture is then taken and poured into the water.
Is there more wine in the water or is there more water in the wine? well, there is more water in wine than wine in water, assumed you wanted to say the glasses poured are the same quantity. + Show Spoiler +the 1st time you take just water, pour into wine the 2nd time you take wine+water and pour back into water
+ Show Spoiler +Your answer is wrong, see posts a few pages back. They are equal. (You're forgetting that in your second step, you're removing some of the water you just put in in the first step. Think about how much total liquid of each type remains, and where it can be.)
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On December 07 2010 04:26 MindRush wrote:Show nested quote +On December 06 2010 12:02 EmptyW wrote:On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler + The first part is simple, as there's no possibility of making it under 7 turns, but there are 7 +6+5+4+3+2+1= 28 paths to do it, depending on the order of your flips. Do consider that if the coins starts as 4 heads/tails, it'll still count as if you've done it. Hence the 7 turns. (To make it more clear, if the coins starts as head, tails, head, tails, and you choose to flip the tails in this setup(whilst not knowing they were tails): the task would be done in the first turn. However, if they started out as head head tails head; it'd set them to head tails tails tails, in which you flip the head for that setup and clear the task in the second turn. Doing this in order so every possibility is given would take 7 turns whatever path you choose to take, as no path would make 2 or more setups to get cleared in one turn.
The thing about the second part is that when he turns the table, the possibilities of your path might or might not change. So out of these 28 paths you have to find the one that gets it right every turn no matter what order the coins are in at the time. Which to me sounds impossible. Therefore I figure it's a trick question, the blind man can tell which coins are heads or tails with his fingertips, so he only needs one turn. this is possible and it goes like this logical data + Show Spoiler +A 2x2 checkers board can take 4 possible states 1: 4 coins same side - called from now on 4C 2: 1 coins same side (doesn't matter which 3 coins) - called from now on 1C 3: 2 adjacent coins same side - called from now on 2A 4: 2 diagonal coins same side - called from now on 2D -the 3 coins same side is the same as 3 coins same side There are couple of moves we are going to analyze: flip a single random coin = F1 flip two adjacent coins = F2A flip two diagonal coins = F2D No matter which position the board is in, F1(1C) = 4C or 2D or 2A F1(2A or 2D) = 1C F2A(1C) = 1C F2A(2A) = 4C or 2D F2A(2D) = 2D F2D(1C) = 1C F2D(2A) = 2A F2D(2D) = 4C Looking at the things above, we see that if we perform the moves F2D, F2A, F2D on a board with 2A or 2D we will eventually get 4C at some point The problem is that the board might be 1C then the solution is simple + Show Spoiler + F2D, F2A, F2D then F1 then F2D, F2A, F2D again - 7 moves in one of these, we are bound to encounter 4C
Thanks. I'd have figured it out myself if I wasn't so sure that it wouldn't work but gj =]
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On December 07 2010 04:43 Pandonetho wrote: Guys I still don't understand the Monty Hall paradox or whatever it's called, and I never have understood it.
If the Host knows that door 3 is a goat and shows it to you, how does it suddenly become a 66% chance of winning if you switch to door 2?
Wouldn't it just become a 50/50 chance to win? Ugh my brain's gonna explode from trying to figure this out. The 100 door example didn't help either, but I sort of understood it when one of the posters said that it would only make sense to stay if you had chosen correctly in the first place. Out of 100 doors if you knew you had a 1% chance to get the right door, and he opened 98 other doors, wouldn't your chance of being correct just increase until it was 50%? + Show Spoiler + Well there are three possible combination for the goats and the car. Say, we pick gate 1.
Gate 1: Car, Gate 2: Goat, Gate 3: Goat - Host randomly opens any of the two gates. We switch -> We LOSE.
Gate 1: Goat, Gate 2: Car, Gate 3: Goat - Host has to open gate 3. We switch -> We WIN.
Gate 1: Goat, Gate 2: Goat, Gate 3: Car - Host has to open gate 2. We switch -> We WIN.
So, if we switch we have a 66% chance to win.
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On December 07 2010 05:57 EmptyW wrote:Show nested quote +On December 07 2010 04:26 MindRush wrote:On December 06 2010 12:02 EmptyW wrote:On December 05 2010 16:31 jeeneeus wrote:The blind man and checker board+ Show Spoiler +This is a two part question. Only open the second spoiler if you are 90% sure you have the correct solution to the first part.A blind man is given a square checkerboard with 4 squares. Each square has a coin on it, and he is told that during each 'turn', he can flip as many coins as he wants. His goal is to make sure that at some point in time, all four of the coins were facing the same way (eg all heads or all tails). He will be given as many of these 'turns' as he wants, but he must try to guarantee a solution in the least amount of turns as possible. How many turns does he need to guarantee that at some point in time, all four coins were facing the same direction? + Show Spoiler +Congrats, now this time, he is given a harder task: in between each turn, a stranger will come and rotate the checkerboard however he wants to. The stranger can decide not to rotate it at all if he wants. The blind man will obviously have no indication of how or whether the board was rotated at all. The blind man is asked to guarantee a solution in the same manner as the first part. Hint:+ Show Spoiler +Part 2 can be completed in the same number of turns as part 1 + Show Spoiler + For part one, I think the answer is 8. There are 4 squares, each with a quarter that can be in one of 2 orientations. That means there are 2^4=16 different possibilities. Now, I had to create a methodical way to get every possible orientation. I find it's easiest if you think of the squares in a row, instead of in a square, so you can think of the orientations as a binary sequence. The original orientation would be written as 0000. Now you flip coins up in a ribbon sequence, meaning you go up to a certain point, then back down, and repeat until you get them all. So the first flip would be 0001. Then 0011 0010 0110 0100 0101 0111 1111 1011 1001 1000 1010 1110 1100 1101. This will cover every possible orientation (as proof, there are 16 of these, when you count 0000). Now you want to find the latest possible one with no earlier inverse (meaning there are 1's where there were 0's and vice versa). For example, the inverse of 1011 is 0100. This is because if there was an earlier inverse, it would have had all four coins with the same side, just the opposite side (heads instead of tails or vice versa). So if the coins were HTHH, although the flip 1011 would make them all Tails, the earlier 0100 would have made them all Heads. The latest of these is 0111, which is 8 total flips. Part two seems trickier, and I don't know if I wanna spend more time on it.
+ Show Spoiler + The first part is simple, as there's no possibility of making it under 7 turns, but there are 7 +6+5+4+3+2+1= 28 paths to do it, depending on the order of your flips. Do consider that if the coins starts as 4 heads/tails, it'll still count as if you've done it. Hence the 7 turns. (To make it more clear, if the coins starts as head, tails, head, tails, and you choose to flip the tails in this setup(whilst not knowing they were tails): the task would be done in the first turn. However, if they started out as head head tails head; it'd set them to head tails tails tails, in which you flip the head for that setup and clear the task in the second turn. Doing this in order so every possibility is given would take 7 turns whatever path you choose to take, as no path would make 2 or more setups to get cleared in one turn.
The thing about the second part is that when he turns the table, the possibilities of your path might or might not change. So out of these 28 paths you have to find the one that gets it right every turn no matter what order the coins are in at the time. Which to me sounds impossible. Therefore I figure it's a trick question, the blind man can tell which coins are heads or tails with his fingertips, so he only needs one turn. this is possible and it goes like this logical data + Show Spoiler +A 2x2 checkers board can take 4 possible states 1: 4 coins same side - called from now on 4C 2: 1 coins same side (doesn't matter which 3 coins) - called from now on 1C 3: 2 adjacent coins same side - called from now on 2A 4: 2 diagonal coins same side - called from now on 2D -the 3 coins same side is the same as 3 coins same side There are couple of moves we are going to analyze: flip a single random coin = F1 flip two adjacent coins = F2A flip two diagonal coins = F2D No matter which position the board is in, F1(1C) = 4C or 2D or 2A F1(2A or 2D) = 1C F2A(1C) = 1C F2A(2A) = 4C or 2D F2A(2D) = 2D F2D(1C) = 1C F2D(2A) = 2A F2D(2D) = 4C Looking at the things above, we see that if we perform the moves F2D, F2A, F2D on a board with 2A or 2D we will eventually get 4C at some point The problem is that the board might be 1C then the solution is simple + Show Spoiler + F2D, F2A, F2D then F1 then F2D, F2A, F2D again - 7 moves in one of these, we are bound to encounter 4C
Thanks. I'd have figured it out myself if I wasn't so sure that it wouldn't work but gj =]
Was going to solve this one, but seems I am too late someone already did . It is somewhat like a rubix-cube in solution
No on has attempted to answer the revised and greatly more difficulty version of the Faustian board-game riddle I posted several pages back. Maybe I made it to difficult( If you do not have a mastery of liberty racing games this may be way to difficult, as searching on Google for this answer will not be easy, so poeple are not going to be able to fake there way through this as well
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Going to re-post this riddle I crafted with a few changes (leaving the answers out , that's right many solutions to this one ), someone attempt it if you dare !!!!! No way to Google this one easily guys as I wrote it , so I imagine very few will attempt it , since cheating is not as practical ( all the riddles anyone can Google an answer to in 2 seconds have been immediately answered .. coincidence , I dont think so)
The Faustian board-game part 2
This is a variant of the " Faustian" board game riddle , that I just came up after reading your riddle in the OP as it reminded me somewhat of a game that already exists(hint) so I will re-list the rules with the changes edited in.....I would say if you dont what game I am basing this on, this will be a very difficult riddle to solve, and even if you do know it could be very difficult still ..GL (however, if you have seen a particular anime you might know the answer right away)
You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a square table . He gives himself and you a huge pile of quarters.The table has 57 by 57 squares on its surface made from what seems to be jade, forming an algebraic pattern with the inlay. He says "ok, Here are the rules : we'll take turns putting quarters down on a square , this equals one turn , or you may pass on any turn . No overlapping allowed, and the quarters must rest in a square, You both then draw lots for whom will go first , and the Devil wins the pick(but little do you know he cheated) . He says with chilling grin " you will be heads I will be tails, heads goes first" . "When a quarter or group of adjacent quarters sharing liberties is surrounded by a quarter or quarters that share liberties, the opponents will be removed." he says with an icy tone . " We can call quarters sharing liberties in some form of chain "a group" ." "The person whom has more quarters captured, as well as squares controlled at the end of the game will win . One may not play in such a way as to recreate the board position following one's previous move. It is also illegal if it would have the effect (after all steps of the play have been completed) of creating a position that has occurred previously in the game, an exception to this is passing . If the table fills up completely , the game ends , whomever has more captured quarters and points wins .A player may not be forced into committing suicide. Liberties that are controlled within groups borders are counted as points at the end of play . " "The game ends if two consecutive passes take place then the following highest score wins. If the the table fills up and no quarters have been captured by either player, the player that played the most moves or has the most points wins, the same applies if there are consecutive passes while squares are still open." (We will use an algebraic grid to name squares, the grid being 57 by 57 and going in the order X,Y )
The devil honorably gives you several moments to contemplate what he just told you and asks if you are ready, after an angst-forged swallow you somberly nod in acquiescence . You both take your seats you on the bottom of the table the devil on the upper end ,and the devil confidently plays the middle square . You nervously play a quarter on the bottom right corners 4' 4' square( meaning 4 squares from the bottom and 4 from the side which would be the 53' 53 point ) . He then mirrors you , by playing the upper left corner at the 4' 4' point . You then pause , knowing something about both mathematics and boardgames you have analyzed his strategy already, by taking the middle square first and mirroring your moves, the devil hopes to avoid any captures and claim victory when the table finally fills up, as he will be the last to play as long as he continues to mirror your play; it is a simple enough strategy.It seems he has plays based on whomever moves first has a forced win using this mirror strategy, as long as tactics are avoided .Then it hit you . You pause deep in thought . EUREKA you have it !!!!!! A lamp hotter than the surface of the sun explodes from your third eye . You have it ... Oh yes you have it...The devil is going to be in for a surprise .....
1. Is there way to win the game ASSUMING the devil CONTINUES in his strategy of mirroring your every move 2 .and if the answer is yes, what is the method used
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On December 05 2010 17:10 SubtleSense wrote: For the 12 balls one that awnser is stupid and done the wrong way, put 6 and 6 one will be heavier, from the heavier group do 3 and 3, and from the heavier group of that one measure 1 on each side if both are equal the one you didnt measure is heavier if 1 is heavier then the other then thats the heavier ball.
Nope. That's what i thought at first, HOWEVER if one ball îs lighter, not heavier it dosen't work. Yo do the first 6/6 split and one group will be higher than the other, yyou do the 3/3 on that one and you fine out that both groups stay the same = a ball îs lighter and therfore in the other pile of 6. But now, you allready used the device for nothing once.
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On December 07 2010 06:28 MrProphylactic wrote: Going to re-post this riddle I crafted with a few changes (leaving the answers out , that's right many solutions to this one ), Pleas someone attempt it if you dare !!!!!
The Faustian board-game part 2
This is a variant of the " Faustian" board game riddle , that I just came up after reading your riddle in the OP as it reminded me somewhat of a game that already exists(hint) so I will re-list the rules with the changes edited in.....I would say if you dont what game I am basing this on, this will be a very difficult riddle to solve, and even if you do know it could be very difficult still ..GL (however, if you have seen a particular anime you might know the answer right away)
You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a square table . He gives himself and you a huge pile of quarters.The table has 57 by 57 squares on its surface made from what seems to be jade, forming an algebraic pattern with the inlay. He says "ok, Here are the rules : we'll take turns putting quarters down on a square , this equals one turn , or you may pass on any turn . No overlapping allowed, and the quarters must rest in a square, You both then draw lots for whom will go first , and the Devil wins the pick(but little do you know he cheated) . He says with chilling grin " you will be heads I will be tails, heads goes first" . "When a quarter or group of adjacent quarters sharing liberties is surrounded by a quarter or quarters that share liberties, the opponents will be removed." he says with an icy tone . " We can call quarters sharing liberties in some form of chain "a group" ." "The person whom has more quarters captured, as well as squares controlled at the end of the game will win . One may not play in such a way as to recreate the board position following one's previous move. It is also illegal if it would have the effect (after all steps of the play have been completed) of creating a position that has occurred previously in the game, an exception to this is passing . If the table fills up completely , the game ends , whomever has more captured quarters and points wins .A player may not be forced into committing suicide. Liberties that are controlled within groups borders are counted as points at the end of play . " "The game ends if two consecutive passes take place then the following highest score wins. If the the table fills up and no quarters have been captured by either player, the player that played the most moves or has the most points wins, the same applies if there are consecutive passes while squares are still open." (We will use an algebraic grid to name squares, the grid being 57 by 57 and going in the order X,Y )
The devil honorably gives you several moments to contemplate what he just told you and asks if you are ready, after an angst-forged swallow you somberly nod in acquiescence . You both take your seats you on the bottom of the table the devil on the upper end ,and the devil confidently plays the middle square . You nervously play a quarter on the bottom right corners 4' 4' square( meaning 4 squares from the bottom and 4 from the side which would be the 53' 53 point ) . He then mirrors you , by playing the upper left corner at the 4' 4' point . You then pause , knowing something about both mathematics and boardgames you have analyzed his strategy already, by taking the middle square first and mirroring your moves, the devil hopes to avoid any captures and claim victory when the table finally fills up, as he will be the last to play as long as he continues to mirror your play; it is a simple enough strategy.It seems he has plays based on whomever moves first has a forced win using this mirror strategy, as long as tactics are avoided .Then it hit you . You pause deep in thought . EUREKA you have it !!!!!! A lamp hotter than the surface of the sun explodes from your third eye . You have it ... Oh yes you have it...The devil is going to be in for a surprise .....
1. Is there way to win the game ASSUMING the devil CONTINUES in his strategy of mirroring your every move 2 .and if the answer is yes, what is the method used
Haha! A GO problem! Sure you can. You just play in contact with his first move (tengen). And using a tatuat simple sequence you can force him in a dango shape with too few liberties, and you just capture his pieces if he just keeps mirroring you. And then using the ponnuki in the center you go on to smash him..
I am allmost shodan btw, PM if you would like to play sometime..
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edit : double posted again sorry , please delete
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On December 07 2010 06:52 Dagon wrote:Show nested quote +On December 07 2010 06:28 MrProphylactic wrote: Going to re-post this riddle I crafted with a few changes (leaving the answers out , that's right many solutions to this one ), Pleas someone attempt it if you dare !!!!!
The Faustian board-game part 2
This is a variant of the " Faustian" board game riddle , that I just came up after reading your riddle in the OP as it reminded me somewhat of a game that already exists(hint) so I will re-list the rules with the changes edited in.....I would say if you dont what game I am basing this on, this will be a very difficult riddle to solve, and even if you do know it could be very difficult still ..GL (however, if you have seen a particular anime you might know the answer right away)
You die and the devil says he'll let you go to heaven if you beat him in a game. the devil sits you down at a square table . He gives himself and you a huge pile of quarters.The table has 57 by 57 squares on its surface made from what seems to be jade, forming an algebraic pattern with the inlay. He says "ok, Here are the rules : we'll take turns putting quarters down on a square , this equals one turn , or you may pass on any turn . No overlapping allowed, and the quarters must rest in a square, You both then draw lots for whom will go first , and the Devil wins the pick(but little do you know he cheated) . He says with chilling grin " you will be heads I will be tails, heads goes first" . "When a quarter or group of adjacent quarters sharing liberties is surrounded by a quarter or quarters that share liberties, the opponents will be removed." he says with an icy tone . " We can call quarters sharing liberties in some form of chain "a group" ." "The person whom has more quarters captured, as well as squares controlled at the end of the game will win . One may not play in such a way as to recreate the board position following one's previous move. It is also illegal if it would have the effect (after all steps of the play have been completed) of creating a position that has occurred previously in the game, an exception to this is passing . If the table fills up completely , the game ends , whomever has more captured quarters and points wins .A player may not be forced into committing suicide. Liberties that are controlled within groups borders are counted as points at the end of play . " "The game ends if two consecutive passes take place then the following highest score wins. If the the table fills up and no quarters have been captured by either player, the player that played the most moves or has the most points wins, the same applies if there are consecutive passes while squares are still open." (We will use an algebraic grid to name squares, the grid being 57 by 57 and going in the order X,Y )
The devil honorably gives you several moments to contemplate what he just told you and asks if you are ready, after an angst-forged swallow you somberly nod in acquiescence . You both take your seats you on the bottom of the table the devil on the upper end ,and the devil confidently plays the middle square . You nervously play a quarter on the bottom right corners 4' 4' square( meaning 4 squares from the bottom and 4 from the side which would be the 53' 53 point ) . He then mirrors you , by playing the upper left corner at the 4' 4' point . You then pause , knowing something about both mathematics and boardgames you have analyzed his strategy already, by taking the middle square first and mirroring your moves, the devil hopes to avoid any captures and claim victory when the table finally fills up, as he will be the last to play as long as he continues to mirror your play; it is a simple enough strategy.It seems he has plays based on whomever moves first has a forced win using this mirror strategy, as long as tactics are avoided .Then it hit you . You pause deep in thought . EUREKA you have it !!!!!! A lamp hotter than the surface of the sun explodes from your third eye . You have it ... Oh yes you have it...The devil is going to be in for a surprise .....
1. Is there way to win the game ASSUMING the devil CONTINUES in his strategy of mirroring your every move 2 .and if the answer is yes, what is the method used
Haha! A GO problem! Sure you can. You just play in contact with his first move (tengen). And using a tatuat simple sequence you can force him in a dango shape with too few liberties, and you just capture his pieces if he just keeps mirroring you. And then using the ponnuki in the center you go on to smash him.. I am allmost shodan btw, PM if you would like to play sometime..
hehe very good , we have a go player .. chess master and shodan here myself , I was hoping you would give non go players a chance ; ) If you play on kgS we can play anytime, (I was about 2 dan on kgs , played for about 8 months last year , but have taken a break and has been longer than that long since I have even played one game . So would need to play a few games before I got that level back, I was studying it quite a bit ,and have forgotten most of what I learned) although i have not played much go this year to be honest I would need to knock the rust off ) As a purely algebraic problem this is interesting imo in itself , but of course any go player would immediately recognize i as a Tengen opening " or mirror go " and of course the anime I was referring to was "Hikaru no go " . When I read the Faustian riddle in the Op that is what it reminded me of , Hence where I got the idea. But yeah, if you know this already it is very simple
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i used to play in orobaduk, but sure we can play in kgs.. I used kgs only to Take some lessons.. I would also like to play some chess but i suck at it
I don't know how can i contact you, though.. Îs there a TL channel on kgs?
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On December 07 2010 07:02 Dagon wrote: i used to play in orobaduk, but sure we can play in kgs.. I used kgs only to Take some lessons.. I would also like to play some chess but i suck at it I don't know how can i contact you, though.. Îs there a TL channel on kgs? We can make our own channels !!!! , lol so we could be the ones to pioneer the tL channel . I haven't even been there for awhile just Pm me here , until we figure it out , I will give you an e-mail as well
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father's will + Show Spoiler + In a father's will he says that he will leave 1/2 of his fortune to hiis eldest son, 1/3 to the middle son, and 1/9 to the youngest son. They split up land and wealth fairly easily, but a problem arises when they try to split up the father's prized stable, that contains 17 horses. 17 does not split evenly among 1/2, 1/3 and 1/9. One son surgests that the remaing horces be cut up, so that each son gets exactly what the father had willed, one son surgests they sell the remainder horses, spliting the money, and the final son surgests that some sort of trade is aranged so that one son will get more horses and less land or something like that. As the sons argue, a wise man rides up and over hears the argument and rides up to the men. With one action, he is able to find a solution that all of the sons are happy with. What did he do?
+ Show Spoiler + The first son gets 9 the second son gets 6, and the third son gets 2.
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Ah, thanks hacklebeast and Ghazwan. That actually makes a lot of sense now!
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Sleeping beauty problem, probability.
+ Show Spoiler +An experiment is conducted with a woman. She knows all the details of the experiment in advance.
On Sunday she is drugged and put to sleep.
While she is asleep a coin is flipped.
If it lands heads then she'll be awoken once on Monday and not drugged again.
If it lands tails then she'll be awoken once on Monday, drugged, put back to sleep and then she'll awake again on Tuesday with no memory of being awoken on Monday.
In essence any time she wakes up she has no way of knowing if it's Monday or Tuesday or the result of the coin flip.
Any time she awakes she is asked "What is your credence now for the proposition that our coin landed heads?"
Answer
+ Show Spoiler +http://en.wikipedia.org/wiki/Sleeping_Beauty_problem#Solutions
On December 06 2010 18:05 Ghazwan wrote:
His solution had an error and was thus proven wrong, or do I remember the wrong guy?
His first proof was flawed but he fixed it and is now credited for proving the theorem. The wonders of wikipedia.
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Conditional, Logical: You have 12 balls which are identical in size and appearance. You know one of the balls is an odd weight (could be either light or heavy).
You have a set scales which can only compare the weight placed on each side. It either reads: Right side is heavier than left side, Left side is heavier than right side or Right side weights the same as left side.
You may only use the scale three times to determine which ball is the odd one and whether it is a heavy or light ball.
How do you go about it?
Hint: + Show Spoiler +There are a total of 24 possible outcomes: 1) A is heavy 2) A is light 3) B is heavy 4) B is light 5) C is heavy ... 24) L is light. So, your solution should be able to cover each outcome regardless of the order in which you use the scale.
Solution:+ Show Spoiler + 1) ABCD = EFGH (Either I, J, K or L is the odd ball) 1.1) IJ = KA 1.1.1) L > A (L is heavy) 1.1.2) L < A (L is light) 1.2) IJ > KA (Either I J are heavy, or K is light) 1.2.1) I > J (I is heavy) 1.2.2) I < J (J is heavy) 1.2.3) I = J (K is light)
1.3) IJ < KA (Either I J are light, or K is heavy) 1.3.1) I > J (I is light) 1.3.2) I < J (J is light) 1.3.3) I = J (K is heavy)
2) ABCD > EFGH 2.1) ABE = CFK (Either D is heavy, or G H is light) 2.1.1) G > H (H is light) 2.1.2) G = H (D is heavy) 2.1.3) G < H (G is light)
2.2) ABE > CFK (Either A B are heavy, or F is light) 2.2.1) A = B (F is light) 2.2.2) A > B (A is heavy) 2.2.3) A < B (B is heavy)
2.3) ABE < CFK (Either C is heavy, or E is light) 2.3.1) C = K (E is light) 2.3.2) C > K (C is heavy)
3) ABCD < EFGH 3.1) ABE = CFK (Either D is light, or G H are heavy) 3.1.1) G > H (G is heavy) 3.1.2) G = H (D light) 3.1.3) G < H (H is heavy)
3.2) ABE < CFK (Either A B are light, or F is heavy) 3.2.1) A = B (F is heavy) 3.2.2) A > B (B is light) 3.2.2) A < B (A is light)
3.3) ABE > CFK (Either E is heavy or C is light) 3.3.1) E = K (C is light) 3.3.2) E > K (E is heavy)
On December 05 2010 13:12 t3tsubo wrote:3 Hats [tag: logic]+ Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be? Solution + Show Spoiler + There are 4 possibilities for B and C to be wearing. 1) B has a white hat, C has a white hat. 2) B has a black hat, C has a white hat. 3) B has a white hat, C has a black hat. 4) B has a black hat, C has a black hat.
Case #1: -There are only 2 white hats. -A can see 2 white hats (B and C) -A would have known his own hat color, but he doesn't know. -Thus: Case#1 is busted.
Case#2: -B sees C's white hat -If B was wearing a white hat, we would have case #1 and A would have known his own hat color, but A didn't know. -B could have used this information to conclude his hat is black. But B didn't know the color of his hat. -Thus: Case#2 is busted.
Case#3 and #4: -Are the only combination where both A and B won't be sure their hat colors. -C is wearing a black hat in both cases. -C ALWAYS "guesses" his hat's color right.
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On December 07 2010 12:54 Leath wrote:Conditional, Logical:You have 12 balls which are identical in size and appearance. You know one of the balls is an odd weight ( could be either light or heavy). You have a set scales which can only compare the weight placed on each side. It either reads: Right side is heavier than left side, Left side is heavier than right side or Right side weights the same as left side. You may only use the scale three times to determine which ball is the odd one and whether it is a heavy or light ball. How do you go about it? Hint: + Show Spoiler +There are a total of 24 possible outcomes: 1) A is heavy 2) A is light 3) B is heavy 4) B is light 5) C is heavy ... 24) L is light. So, your solution should be able to cover each outcome regardless of the order in which you use the scale. Solution:+ Show Spoiler + 1) ABCD = EFGH (Either I, J, K or L is the odd ball) 1.1) IJ = KA 1.1.1) L > A (L is heavy) 1.1.2) L < A (L is light) 1.2) IJ > KA (Either I J are heavy, or K is light) 1.2.1) I > J (I is heavy) 1.2.2) I < J (J is heavy) 1.2.3) I = J (K is light)
1.3) IJ < KA (Either I J are light, or K is heavy) 1.3.1) I > J (I is light) 1.3.2) I < J (J is light) 1.3.3) I = J (K is heavy)
2) ABCD > EFGH 2.1) ABE = CFK (Either D is heavy, or G H is light) 2.1.1) G > H (H is light) 2.1.2) G = H (D is heavy) 2.1.3) G < H (G is light)
2.2) ABE > CFK (Either A B are heavy, or F is light) 2.2.1) A = B (F is light) 2.2.2) A > B (A is heavy) 2.2.3) A < B (B is heavy)
2.3) ABE < CFK (Either C is heavy, or E is light) 2.3.1) C = K (E is light) 2.3.2) C > K (C is heavy)
3) ABCD < EFGH 3.1) ABE = CFK (Either D is light, or G H are heavy) 3.1.1) G > H (G is heavy) 3.1.2) G = H (D light) 3.1.3) G < H (H is heavy)
3.2) ABE < CFK (Either A B are light, or F is heavy) 3.2.1) A = B (F is heavy) 3.2.2) A > B (B is light) 3.2.2) A < B (A is light)
3.3) ABE > CFK (Either E is heavy or C is light) 3.3.1) E = K (C is light) 3.3.2) E > K (E is heavy)
A fan of dinosaur comics, I'm guessing?
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On December 07 2010 02:08 theSAiNT wrote:Show nested quote +On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. The problem is posted wrongly. You said the fire moves from A to B and the wind moves from A to B. In which case there is no solution. + Show Spoiler + The answer was posted in the thread. You light a fire farther up and stand in between the two, then stand in the ashes of the one you lit.
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On December 07 2010 04:43 Pandonetho wrote: Guys I still don't understand the Monty Hall paradox or whatever it's called, and I never have understood it.
If the Host knows that door 3 is a goat and shows it to you, how does it suddenly become a 66% chance of winning if you switch to door 2?
Wouldn't it just become a 50/50 chance to win? Ugh my brain's gonna explode from trying to figure this out. The 100 door example didn't help either, but I sort of understood it when one of the posters said that it would only make sense to stay if you had chosen correctly in the first place. Out of 100 doors if you knew you had a 1% chance to get the right door, and he opened 98 other doors, wouldn't your chance of being correct just increase until it was 50%?
It works because the host isn't opening a door at random. If you have 3 doors and the host randomly opens up door 2 or 3 and it can contain the car or the goat and it just happens to contain the goat, then it won't benefit you to switch. The reason it works is because the host is not picking a door to open at random, he is specifically picking a door that does not contain the prize.
If there are 1 million doors and you pick door #24, and the host opens every door except for door #494,296, and they all contain goats, then you probably don't have a 50/50 chance of winning if you stay with your own door. It's not like he randomly decided to skip door #494,296 and open a million other doors. There is a very good chance that is the door with the car.
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I finally solved the 12 balls without looking at any answers! I am excited! Even though the solution is posted think I shall type mine. + Show Spoiler + Measure 4 of them against another 4. If they are equal, those 8 are normal we must now consider the last 4 that we have no information about:
Measure 3 of the remaining ones against any 3 of the 8 you now know are normal. If they are equal we know the last one is odd we just need to know its weight: Measure the last 1 against any other to tell if it's too heavy or too light. If they are heavier: weigh two of them against each other, and if they are equal, the third is too heavy. If they are unequal, the heavier one is too heavy. Ditto if they are too light, just switch heavy with light.
If the first 4v4 is unequal we can split the balls into a "heavy group" and a "light group" meaning if one of them is the odd ball, we know it is odd in the direction of that group: On one side measure one ball from the heavy group, one ball from the light group and one of the 4 remaining ones you know are normal. On the other side place two balls from the light group and one from the heavy group.
If they are equal: Measure the two heavy ones you left out against each other. If they are equal, the third (light group left out) is too light. If they are unequal, the heavier one is too heavy.
If the 2nd 3v3 is unequal: If the heavy/light/normal group is heavier, measure the two light balls from the heavy/light/light group against each other, if they are unequal, the lighter one is too light, if they are equal, the one that came from the very first heavy group on the heavy/light/normal side is too heavy. If the heavy/light/light group is heavier, measure the heavy ball on this side against a normal ball, if it's unequal, that heavy one is too heavy, if they are equal, the ball from the initial light group on the other side is too light.
Tada!
Looking at the posted solution my first measurement is equal method is slightly different but it still works.
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hello there, not advertising or anything, don't have any interest of you buying this game or not, but, really checkout this -> http://braid-game.com/ if you like logic games and riddles, you will LOVE Braid.
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On December 07 2010 14:20 Trion wrote:Show nested quote +On December 07 2010 02:08 theSAiNT wrote:On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. The problem is posted wrongly. You said the fire moves from A to B and the wind moves from A to B. In which case there is no solution. + Show Spoiler + The answer was posted in the thread. You light a fire farther up and stand in between the two, then stand in the ashes of the one you lit.
This entire line of thought is illogical.
Let us see why: 1) Fire going from A to B at speed 1 mph. 2) Wind blowing from B to A at 2 mph (assuming this was a typing mistake)
Your answer: + Show Spoiler + Start a fire in the opposite direction, a mid point C. The new fire at C, with the help of the wind will burn from C towards A until it meets the original fire. By then, there will be only ashes.
It does not make sense, because... + Show Spoiler + You must expect both fires to behave the same. If the initial fire was burning from A to B against the wind, why would a mid point fire not carry on from C to B as well at the same rate as the first one? Willywuntang should probably just try digging a ditch in a strategical space where trees and fire are least likely to get inside, curl into a ball and hope for the best. Lol.
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Posted this on Page 4, no replies?
Two Envelopes Say you're given two envelopes, each one with a real amount of money in it. You have no clue how much is in each, but the only information that's given about it is that one contains double the amount of the other. You have to pick one, you can open it, look at the amount, and you can decide to then switch and pick the other.
Now explain this paradox: Just picking one (which would give you amount X) is not as good as picking one first, then switching to the other envelope, which gives you (0.5 * X * 0.5 + 0.5 * X * 2) / 2, which is 1.25 * X on average. But that means you could just divert from your initial choice and get more? Please explain!
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11 pages buried deep, i found this.
On December 05 2010 14:13 Aeres wrote:I'm still trying to figure this one out (from Ares[Effort]'s riddle thread): Show nested quote +On March 16 2009 14:27 l10f wrote: The MBC Game Hero team decided to go on a trip to the USA. Light decided to bring his laptop to practice with him. All the other players were jealous that he could practice while they couldn't. One day, his laptop went missing while he slept.
1. Jaehoon claims that he was busy hooking up with girls last night. 2. Shark says that he never even saw the laptop. 3. Pusan says he saw Sea take the laptop. 4. Sea says he was out last night on a one night stand with a girl he met, and that Pusan is lying 5. Saint says Shark was in Light's room at 3AM last night.
Any of the players could be telling a lie, or could be telling the truth. Who stole Light's laptop?
I just wonder.. Why did Shark say he never even saw the laptop? He was supposedly jealous of it according to the intro.
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On December 07 2010 20:15 Leath wrote:Show nested quote +On December 07 2010 14:20 Trion wrote:On December 07 2010 02:08 theSAiNT wrote:On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. The problem is posted wrongly. You said the fire moves from A to B and the wind moves from A to B. In which case there is no solution. + Show Spoiler + The answer was posted in the thread. You light a fire farther up and stand in between the two, then stand in the ashes of the one you lit.
This entire line of thought is illogical. Let us see why: 1) Fire going from A to B at speed 1 mph. 2) Wind blowing from B to A at 2 mph (assuming this was a typing mistake) Your answer: + Show Spoiler + Start a fire in the opposite direction, a mid point C. The new fire at C, with the help of the wind will burn from C towards A until it meets the original fire. By then, there will be only ashes.
It does not make sense, because... + Show Spoiler + You must expect both fires to behave the same. If the initial fire was burning from A to B against the wind, why would a mid point fire not carry on from C to B as well at the same rate as the first one? Willywuntang should probably just try digging a ditch in a strategical space where trees and fire are least likely to get inside, curl into a ball and hope for the best. Lol.
+ Show Spoiler +No matter which way the second fire burns, it'll leave a trail of ash which the first fire won't be able to get through. You have to remember that it'll take some time for the first fire to get from A to C and in this time the second fire will have progressed from C in the direction of B and thus the first fire won't have anything to burn. As long as willywutang stands in the ashes left by the second fire he will be fine. The only problem will be his survival afterwards seeing as the entire island will be burnt down
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On December 07 2010 20:15 Leath wrote:Show nested quote +On December 07 2010 14:20 Trion wrote:On December 07 2010 02:08 theSAiNT wrote:On December 05 2010 13:12 t3tsubo wrote:Willywutang and the burning island of doom [tag: logic, thinkoutsidethebox]+ Show Spoiler +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) Answer: + Show Spoiler +Start a fire farther down, the wind will move it in the opposite direction, so when the 2 flames meet they will have no more fuel to burn. The problem is posted wrongly. You said the fire moves from A to B and the wind moves from A to B. In which case there is no solution. + Show Spoiler + The answer was posted in the thread. You light a fire farther up and stand in between the two, then stand in the ashes of the one you lit.
This entire line of thought is illogical. Let us see why: 1) Fire going from A to B at speed 1 mph. 2) Wind blowing from B to A at 2 mph (assuming this was a typing mistake) Your answer: + Show Spoiler + Start a fire in the opposite direction, a mid point C. The new fire at C, with the help of the wind will burn from C towards A until it meets the original fire. By then, there will be only ashes.
It does not make sense, because... + Show Spoiler + You must expect both fires to behave the same. If the initial fire was burning from A to B against the wind, why would a mid point fire not carry on from C to B as well at the same rate as the first one? Willywuntang should probably just try digging a ditch in a strategical space where trees and fire are least likely to get inside, curl into a ball and hope for the best. Lol.
+ Show Spoiler +He'll be dead anyway since burned all vegetation on the island. He just can wait to starve.
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On December 07 2010 04:43 Pandonetho wrote: Guys I still don't understand the Monty Hall paradox or whatever it's called, and I never have understood it.
If the Host knows that door 3 is a goat and shows it to you, how does it suddenly become a 66% chance of winning if you switch to door 2?
Wouldn't it just become a 50/50 chance to win? Ugh my brain's gonna explode from trying to figure this out. The 100 door example didn't help either, but I sort of understood it when one of the posters said that it would only make sense to stay if you had chosen correctly in the first place. Out of 100 doors if you knew you had a 1% chance to get the right door, and he opened 98 other doors, wouldn't your chance of being correct just increase until it was 50%?
Try approaching this problem from the host's perspective. Let's say we have 100 doors. The host knows that he has to open 98 doors with goats after the player chooses a door. Let's say door #1 has the prize. 99% of the time, the host will open all the doors except for door #1 and the door that was chosen by the player. This means that in 99% of the time, the remaining closed door has the prize. Only in 1% of the time is door #1 chosen. And this is the only time that the remaining closed door has a goat.
Does that explain it to you?
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3 Hats [tag: logic] + Show Spoiler +There are 3 black hats and 2 white hats in a box. Three men (we will call them A, B, & C) each reach into the box and place one of the hats on his own head. They cannot see what color hat they have chosen. The men are situated in a way that A can see the hats on B & C's heads, B can only see the hat on C's head and C cannot see any hats. When A is asked if he knows the color of the hat he is wearing, he says no. When B is asked if he knows the color of the hat he is wearing he says no. When C is asked if he knows the color of the hat he is wearing he says yes and he is correct. What color hat and how can this be?
Not sure if anyone's posted an answer to this already but here's mine:
+ Show Spoiler +C's hat is black.
For A to not know what color his hat is, there are one of two options for the colors of B and C's hats.
1 white 1 black, or 2 black.
From B's point of view, he knows it's one of these two options. If C had a white hat, then that means B should know his hat is white.
Since B doesn't know his hat color, the only remaining option is that B and C both have black hats, and C doesn't need to see any colors to know.
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Here are a couple of logic problems.
1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers.
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On December 08 2010 03:56 MindRush wrote:Here are a couple of logic problems. + Show Spoiler + 1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers.
+ Show Spoiler +1. He has the hiccups! 2. They are getting a divorce? 3. His wife is a terrorist.. lol
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United States4053 Posts
On December 08 2010 03:56 MindRush wrote: Here are a couple of logic problems.
1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers. + Show Spoiler +#3 sounds reminiscent of a story from Greek mythology
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A black horse jumps over a castle and a man disappears. Why?
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On December 07 2010 20:57 aseq wrote:
Posted this on Page 4, no replies?
Two Envelopes Say you're given two envelopes, each one with a real amount of money in it. You have no clue how much is in each, but the only information that's given about it is that one contains double the amount of the other. You have to pick one, you can open it, look at the amount, and you can decide to then switch and pick the other.
Now explain this paradox: Just picking one (which would give you amount X) is not as good as picking one first, then switching to the other envelope, which gives you (0.5 * X * 0.5 + 0.5 * X * 2) / 2, which is 1.25 * X on average. But that means you could just divert from your initial choice and get more? Please explain!
Despite your poorly worded riddle, I shall again, answer. + Show Spoiler + The actual problem is called the two envelopes problem (I like to read up on paradoxes in wikipedia in my spare time, they amuse me). Supposing that two men are given envelopes with unknown sums of money and are told only that one envelope contains twice the amount of the other, they are allowed to switch envelopes should they both agree. Each man reasons that if he gets twice the amount, he will have increased his money by 100% and if he gets half the amount, he will only lose 50%. Each man will reason it is beneficial to trade, but upon doing so, makes the same argument again for swapping back, making it an endless loop of swapping. How is it possible for both to benefit from swapping?
Of course simple math solves this riddle easily. This is yet another example of how (and I've been saying this forever) probability is determined by perspective. If there were an omnipresent being that knew what the sums were, surely he would know whether to switch or keep his envelope, so the probability becomes 100/0. However, for someone who does not know, the probability is 50/50. Therein lies the paradox. You assume that there is an infinite loop due to the causality in the nature of the problem. However, it is necessary to define limits at the beginning of the problem. All in all, there is only 3X in money, one envelope with X and another with 2X. The average of these is 1.5X so at the start, you should feel as if you have the average of both envelopes. If you switch, you stand to gain/lose 0.5X, which is equal on both ends. So by taking a 3rd party view on the situation, it becomes clear that neither is actually benefitted by swapping. It is just an illusion created by the perspective of the individual.
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On December 08 2010 12:50 ieatpasta wrote: A black horse jumps over a castle and a man disappears. Why? + Show Spoiler +
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The circus night watchman [tag: logic] + Show Spoiler +There is a night watchman that has worked in a circus for many years now, not any problem whatsoever. One day, the circus owner tells everyone he is going to fly to another country in order to make deals for presentations. The next day, the watchman, frightened, tells the circus owner that he cannot get on the plane, because it is going to explode and he is going to die in the accident. When the circus owner asks how he knows that, the watchmen responds: "I dreamt of it, the explosion, the accident, the funeral... everything. The whole circus suffers from your loss. It was so vivid, please you cannot get into that plane". The circus owner follows the advice and does not go. Later on the news, they know that the plain indeed exploded; no survivors. Finally, the circus owner says to the watchman: "Thank you for saving my life, but I'm going to have to fire you". Question: Why does the circus owner fire the watchmen, after he saved his life? Answer: + Show Spoiler +The job of a watchman is to look after the circus at night, so he has to be always alert. He dreamt that his boss was going to die in the plane, so he was sleeping during work.
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If eggs are 10 cents a dozen and a china man is 10 feet tall, how many pancakes does it take to shingle a dog house?
Answer: + Show Spoiler +None, because ice cream has no bones
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I dont know if the 3 hats one is answer yet so ill just post this and op can update it into the quesiton + Show Spoiler + A can see both B and C. If B is wearing a white hat and C is wearing a white hat, then A is wearing a black hat. He cannot answer so that means B and C are both not wearing white hats. B knows what A knows from his answer. Therefore, he knows that if C is wearing a white hat, B will be wearing a black hat. He cannot answer so therefore C is not wearing a white hat. He is wearing a black hat and his hat could be white or black. C knows what B and A know so he knows hes wearing a black hat.
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On December 08 2010 03:56 MindRush wrote: Here are a couple of logic problems.
1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers.
Here are the answers i promised + Show Spoiler + 1. The guy had hiccups. He wanted water to get rid of them, bartender scared him. 2. The guy and his wife work in a circus. He throws the knives, his wife is on the wheel. She sees him having eye problems, she quits her job. 3. The rich guy was framed by his wife. She staged her murder and put the blame on him. He stood in prison for 40 years for killing his wife. When he got out, he wanted revenge. He killed his wife no matter what, then he turned himself over to the police. But he did the time already for killing his wife in the 1st place, so he gets away with it.
Some other logical questions: 4. Guy was found in an empty room and he was found hanged by a hook in the ceiling. The room was locked from the inside, no sign of lockpicking or anything, so the guy clearly committed suicide. The room is 4m high. The length of the rope is 1m. Police found water on the floor. How did the guy manage to commit suicide? (room completely empty, 4m high, 1m rope, definitely a suicide) 5. Why is 5 afraid of 7 ? 6. What did the big chimney say to the little chimney? 7. What did the big tomato say to the little tomato when they were rolling like banelings ?
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On December 09 2010 08:48 MindRush wrote:Show nested quote +On December 08 2010 03:56 MindRush wrote: Here are a couple of logic problems.
1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers. Here are the answers i promised + Show Spoiler + 1. The guy had hiccups. He wanted water to get rid of them, bartender scared him. 2. The guy and his wife work in a circus. He throws the knives, his wife is on the wheel. She sees him having eye problems, she quits her job. 3. The rich guy was framed by his wife. She staged her murder and put the blame on him. He stood in prison for 40 years for killing his wife. When he got out, he wanted revenge. He killed his wife no matter what, then he turned himself over to the police. But he did the time already for killing his wife in the 1st place, so he gets away with it.
Some other logical questions: 4. Guy was found in an empty room and he was found hanged by a hook in the ceiling. The room was locked from the inside, no sign of lockpicking or anything, so the guy clearly committed suicide. The room is 4m high. The length of the rope is 1m. Police found water on the floor. How did the guy manage to commit suicide? (room completely empty, 4m high, 1m rope, definitely a suicide) 5. Why is 5 afraid of 7 ? 6. What did the big chimney say to the little chimney? 7. What did the big tomato say to the little tomato when they were rolling like banelings ?
Most of those are not actually riddles. They are a detective game, where the host tells the end of the story and all other players have to ask questions, which may only be answered with either yes, no or irrelevant, until they figure out how the story starts. It is impossible to guess the beginning of those stories by the end alone.
"A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why?" Could have any open beginning whatsoever. He could be a spy; the whole water, gun, thanks were just codes. Maybe he was a deadbeat who had a bad habit of not paying his bills, after refusing to pay for the glass of water the bartender points the gun at him, he thanks for helping him become a better and honest person. o_O Anything is possible.
As for the others 5. + Show Spoiler +Because 7 8 (ate) 9. Hehe cute 6. No idea 7. No idea :p
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On December 09 2010 09:17 Leath wrote:Show nested quote +On December 09 2010 08:48 MindRush wrote:On December 08 2010 03:56 MindRush wrote: Here are a couple of logic problems.
1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers. Here are the answers i promised + Show Spoiler + 1. The guy had hiccups. He wanted water to get rid of them, bartender scared him. 2. The guy and his wife work in a circus. He throws the knives, his wife is on the wheel. She sees him having eye problems, she quits her job. 3. The rich guy was framed by his wife. She staged her murder and put the blame on him. He stood in prison for 40 years for killing his wife. When he got out, he wanted revenge. He killed his wife no matter what, then he turned himself over to the police. But he did the time already for killing his wife in the 1st place, so he gets away with it.
Some other logical questions: 4. Guy was found in an empty room and he was found hanged by a hook in the ceiling. The room was locked from the inside, no sign of lockpicking or anything, so the guy clearly committed suicide. The room is 4m high. The length of the rope is 1m. Police found water on the floor. How did the guy manage to commit suicide? (room completely empty, 4m high, 1m rope, definitely a suicide) 5. Why is 5 afraid of 7 ? 6. What did the big chimney say to the little chimney? 7. What did the big tomato say to the little tomato when they were rolling like banelings ? Most of those are not actually riddles. They are a detective game, where the host tells the end of the story and all other players have to ask questions, which may only be answered with either yes, no or irrelevant, until they figure out how the story starts. It is impossible to guess the beginning of those stories by the end alone. "A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why?"Could have any open beginning whatsoever. He could be a spy; the whole water, gun, thanks were just codes. Maybe he was a deadbeat who had a bad habit of not paying his bills, after refusing to pay for the glass of water the bartender points the gun at him, he thanks for helping him become a better and honest person. o_O Anything is possible. As for the others 5. + Show Spoiler +Because 7 8 (ate) 9. Hehe cute 6. No idea 7. No idea :p
this guy got the 1st answer.
On December 08 2010 12:43 ieatpasta wrote:Show nested quote +On December 08 2010 03:56 MindRush wrote:Here are a couple of logic problems. + Show Spoiler + 1. A guy walks into a bar, asks for a glass of water. The bartender pulls a gun at the guy. The guy thanks him. Why ? 2. A guy goes to an eye doctor (ophthalmologist). His wife, who works in the same place as her husband, quits her job the next day. Why ? 3. A rich guy comes back from a place where he has been for 40 years. He goes to his castle, finds and kills his wife. He goes to the nearest police station and tells them he killed his wife. The police let him go, no questions asked, no bribes, no fancy stuff. Why ?
All questions are reasonable and have logic answers. I will come later with answers in spoilers.
+ Show Spoiler +1. He has the hiccups! 2. They are getting a divorce? 3. His wife is a terrorist.. lol
as for 5,6 and 7, + Show Spoiler + 5. Why is 5 afraid of 7 ? because 7 8(ate) 9 6. What did the big chimney say to the little chimney? don't smoke 7. What did the big tomato say to the little tomato when they were rolling like banelings ? catch up
Oki, please somebody post some new shit
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For 4:
+ Show Spoiler +He died standing on an iceblock. Once it melted, he lost his footing and died.
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On December 09 2010 09:46 Antifate wrote:For 4: + Show Spoiler +He died standing on an iceblock. Once it melted, he lost his footing and died.
yeah, thanks. forgot 2 post answer for question 4
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On December 06 2010 08:39 Thoreezhea wrote: A cop is at an intersection on a motorcycle during a red light and 4 teenagers in a car race past him through the intersection at 50 mph. the cop did not attempt to detain them or pursue them in any way. Why?
+ Show Spoiler +Car with teenagers was on the road that had a green light. Road the teenagers are on has a speed limit of 50 (or higher).
Alternately
+ Show Spoiler +If they are on the same road as the cop, the cop was going to turn and had a red arrow while it was green to go straight. Still not breaking the speed limit.
How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy) + Show Spoiler +One of either 1 or 19 could be a negative number while the other is a positive number.
Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours)
I didn't really get the answer mentioned in the first post. Another possibility, with a few assumptions:
+ Show Spoiler +If the fire doesn't linger and burns out as it goes, he could start a fire partway between A and B (point C) as mentioned. The wind would blow the new fire towards B as well. By the time the original fire gets to Point C, the new fire will have moved towards B and burned itself out at C, allowing him to follow the second fire across the ashes. The fire starting at point A will have no fuel to continue and will progress no further and burn itself out. The second Fire will get to the end of the island and burn itself out as well, saving him from a death by burning (but screwing him in other ways if he doesn't get rescued soon, heh).
4. Guy was found in an empty room and he was found hanged by a hook in the ceiling. The room was locked from the inside, no sign of lockpicking or anything, so the guy clearly committed suicide. The room is 4m high. The length of the rope is 1m. Police found water on the floor. How did the guy manage to commit suicide? (room completely empty, 4m high, 1m rope, definitely a suicide)
I heard a slightly fancier version of that one. No water in the room, but the bottom of the man's feet are heavily discolored. Rest of the riddle is the same.
+ Show Spoiler +Stood on a chunk of dry ice, which evaporates and doesn't leave a pool of water. My memory told me that the man's feet are supposed to be burned, but a check on wiki says that prolonged exposure to dry ice causes frostbite... w/e, discoloration works both ways.
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On December 09 2010 02:57 LeGo_MaN wrote:The circus night watchman [tag: logic]+ Show Spoiler +There is a night watchman that has worked in a circus for many years now, not any problem whatsoever. One day, the circus owner tells everyone he is going to fly to another country in order to make deals for presentations. The next day, the watchman, frightened, tells the circus owner that he cannot get on the plane, because it is going to explode and he is going to die in the accident. When the circus owner asks how he knows that, the watchmen responds: "I dreamt of it, the explosion, the accident, the funeral... everything. The whole circus suffers from your loss. It was so vivid, please you cannot get into that plane". The circus owner follows the advice and does not go. Later on the news, they know that the plain indeed exploded; no survivors. Finally, the circus owner says to the watchman: "Thank you for saving my life, but I'm going to have to fire you". Question: Why does the circus owner fire the watchmen, after he saved his life? Answer: + Show Spoiler +The job of a watchman is to look after the circus at night, so he has to be always alert. He dreamt that his boss was going to die in the plane, so he was sleeping during work.
I don't get it. So normally he would sleep during the day if his job as at night right? Why can't he have 'dreamt' of it during the day? Does he just not sleep at all? ^o) *confused*
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On December 09 2010 16:57 Vortok wrote:Show nested quote +On December 06 2010 08:39 Thoreezhea wrote: A cop is at an intersection on a motorcycle during a red light and 4 teenagers in a car race past him through the intersection at 50 mph. the cop did not attempt to detain them or pursue them in any way. Why?
+ Show Spoiler +Car with teenagers was on the road that had a green light. Road the teenagers are on has a speed limit of 50 (or higher). Alternately + Show Spoiler +If they are on the same road as the cop, the cop was going to turn and had a red arrow while it was green to go straight. Still not breaking the speed limit. Show nested quote +How do you take 1 away from 19 and get left with 20 . Two possible answers(difficulty easy) + Show Spoiler +One of either 1 or 19 could be a negative number while the other is a positive number. Show nested quote +Willywutang is hanging out on a heavily forested island that's really narrow: it's a narrow strip of land that's ten miles long. let's label one end of the strip A, and the other end B. a fire has started at A, and the fire is moving toward B at the rate of 1 mph. at the same time, there's a 2 mph wind blowing in the direction from A toward B. what can Willywutang do to save himself from burning to death?! assume that Willywutang can't swim and there are no boats, jetcopters, teleportation devices, etc.. (if he does nothing, Willywutang will be toast after at most 10 hours, since 10 miles / 1 mph = 10 hours) I didn't really get the answer mentioned in the first post. Another possibility, with a few assumptions: + Show Spoiler +If the fire doesn't linger and burns out as it goes, he could start a fire partway between A and B (point C) as mentioned. The wind would blow the new fire towards B as well. By the time the original fire gets to Point C, the new fire will have moved towards B and burned itself out at C, allowing him to follow the second fire across the ashes. The fire starting at point A will have no fuel to continue and will progress no further and burn itself out. The second Fire will get to the end of the island and burn itself out as well, saving him from a death by burning (but screwing him in other ways if he doesn't get rescued soon, heh). Show nested quote +4. Guy was found in an empty room and he was found hanged by a hook in the ceiling. The room was locked from the inside, no sign of lockpicking or anything, so the guy clearly committed suicide. The room is 4m high. The length of the rope is 1m. Police found water on the floor. How did the guy manage to commit suicide? (room completely empty, 4m high, 1m rope, definitely a suicide) I heard a slightly fancier version of that one. No water in the room, but the bottom of the man's feet are heavily discolored. Rest of the riddle is the same. + Show Spoiler +Stood on a chunk of dry ice, which evaporates and doesn't leave a pool of water. My memory told me that the man's feet are supposed to be burned, but a check on wiki says that prolonged exposure to dry ice causes frostbite... w/e, discoloration works both ways.
my way of seeing this + Show Spoiler + 1. cop could be off-duty 2. take I from XIX, you get lest with XX - roman numbers
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On December 07 2010 20:57 aseq wrote:
Posted this on Page 4, no replies?
Two Envelopes Say you're given two envelopes, each one with a real amount of money in it. You have no clue how much is in each, but the only information that's given about it is that one contains double the amount of the other. You have to pick one, you can open it, look at the amount, and you can decide to then switch and pick the other.
Now explain this paradox: Just picking one (which would give you amount X) is not as good as picking one first, then switching to the other envelope, which gives you (0.5 * X * 0.5 + 0.5 * X * 2) / 2, which is 1.25 * X on average. But that means you could just divert from your initial choice and get more? Please explain!
You're wrong that X is your expected value for your first choice. If y is the amount in the bad envelope, you expect to see (y+2y)/2 dollars in either envelope.
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On December 06 2010 11:28 Trion wrote: Alright I bet no one will get this one.
Prove that there is no solution for x^n+y^n=z^n when n>2 and all numbers are natural.
One of the number theorists at my university has this; it's a T-shirt. The above pictures are the front and back. AWESOME.
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why would a man call a phone, when he has a perfectly good and working and accurate watche and a equally functional clock?
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On December 21 2010 10:35 Thoreezhea wrote: why would a man call a phone, when he has a perfectly good and working and accurate watche and a equally functional clock?
He wishes to order a Pizza?
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Figured I'd wake this thread up with a few deceptively easy ones:
1. 10 tigers come across a dead deer in the wild. They are unwilling to share the meal, so the meal would go to any one (the strongest) among them. However, if a tiger eats the deer, that tiger will suffer from food coma and is liable to get eaten by any one among the remaining tigers. That tiger, in turn, will suffer from food coma and is liable to get eaten by any one of the remaining tigers, and so on. Will any tiger eat the deer?
2. There's a game I don't know the name of, but it involves stones (or pebbles) placed in rows and columns like this:
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
There could be as many (whole) rows and columns as you'd like. The game has two players who take turns picking stones to remove. The stone they pick and *all* stones over and to the right of it will be removed, including any stones diagonally up and to the right of it. So if one picks the stone marked with a plus below, the stones marked with a dash will *also* be removed. The goal of the game is to avoid picking the last stone. Whoever has to pick that stone has lost.
. . . - - . . . - - . . . + - . . . . . . . . . . . . . . .
Who will win? The player who goes first, or the player who goes second? (Assume they play perfect).
Note that this is deceptively easy. You don't have to go through any trial-and-error or anything of the sorts, just come to a rather simple realization.
Good luck!
edit: Both solved by qrs (just browse through his posts on this page if you want the answer).
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On January 25 2011 16:47 Mayfly wrote:Figured I'd wake this thread up with a few deceptively easy ones: 1. 10 tigers come across a dead deer in the wild. They are unwilling to share the meal, so the meal would go to any one (the strongest) among them. However, if a tiger eats the deer, that tiger will suffer from food coma and is liable to get eaten by any one among the remaining tigers. That tiger, in turn, will suffer from food coma and is liable to get eaten by any one of the remaining tigers, and so on. Will any tiger eat the deer? 2. There's a game I don't know the name of, but it involves stones (or pebbles) placed in rows and columns like this: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . There could be as many (whole) rows and columns as you'd like. The game has two players who take turns picking stones to remove. The stone they pick and *all* stones over and to the right of it will be removed, including any stones diagonally up and to the right of it. So if one picks the stone marked with a plus below, the stones marked with a dash will *also* be removed. The goal of the game is to avoid picking the last stone. Whoever has to pick that stone has lost. . . . - - . . . - - . . . + - . . . . . . . . . . . . . . . Who will win? The player who goes first, or the player who goes second? (Assume they play perfect). Note that this is deceptively easy. You don't have to go through any trial-and-error or anything of the sorts, just come to a rather simple realization. Good luck!
1. Yes. Assuming all the tigers are rational, if I were one of the tigers, I would eat a deer. I could sleep in peace knowing if any of the tigers chose to eat me, he too would die. Because of this, no tiger would dare eat me.
2. The player who goes first. On your first move you pick this pebble: . - - - - . - - - - . - - - - . x - - - . . . . . Then regardless of what the other player does, just mirror it, and you can't lose.
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Both answers wrong. You're on the right track, though. However, if you were to pick the stone you said, you would definitely lose to a perfect player. You're thinking too much about which stone to pick, it's more about who will win.
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What move set could a perfect player do that would allow him to win from that spot.
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+ . . . . . . . . .
edit: That's why I made it asymmetrical. If it was symmetrical (i.e. a perfect square) it would've been a brilliant move on your part. But as I said, it could be any amount of rows and columns, and I'm looking for a general proof that one of the two players can force a win.
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Oh, ok. I didn't realize it could be any number of rows or columns.
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For the tigers: + Show Spoiler +If there is one waking tiger left + food, of course he eats the food. If there are two waking tigers left + food, neither tiger dares eat the food because he will be eaten after he falls asleep. Therefore if there are three tigers left + food, one tiger can safely eat the food. Therefore if there are four tigers left, no tiger can safely turn himself into food. etc.: if there are an even number of tigers left, no one eats. 10 is an even number. For the stones: + Show Spoiler +If it's true that the game is either always won by the first player or always won by the second player, then it must be always won by the first player. Proof: suppose it were always won by the second player. Then on his first move, the first player could just slice off a row or column to reach a different version of the game where he is the second player, and thus the winner. This leads to a contradiction, so the game must be always won by the first player.
However, this doesn't prove that the game is always won by the first player--maybe some versions of the game are won by the first player and some by the second.
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Correct, dear sir! (edit: for the tigers).
The stones still need some work but you're very much on the right track. (Your proof is actually quite valid, but can be made even more general).
edit 2: Some hints: The game is won by player number + Show Spoiler +. You need to already know that this game (and every game that has no ties) has a means to win it by perfect play, since it has no ties. It's not that proof I'm looking for here.
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LOL damnit i love this, but can someone explain to me something? the one about the monks..ive read the answers and stuff and apparently most of u say if no1 kills themselves within the 1st night it means 2 or more have red eyes...if no1 kills themselves within the 2nd night that means 3 or more have red eyes...etc, can u guys explain this? i dont understand >_< thanks
TL;DR: why, if no monk commits suicide on night 1, means that 2+ monks have red eyes?
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Some riddles: Horses and Tracks + Show Spoiler +You have 25 horses, and each one runs at a unique, constant speed. You have exactly one, straight race track, and it has five lanes, each lane only being able to fit one horse. You have no stopwatch, so the only way to determine a horse's speed is to race it directly against another horse, and see which one is faster. You can put five horses on the track, have them all go, and then record the order of the five. That counts as exactly one race. Your job is to find the three fastest horses in as few races as possible Answer: + Show Spoiler +You can find the three fastest horses in 7 races. First, split the horses into five groups, 1, 2, 3, 4, and 5. Race each of the groups, and then give the horses themselves identifiers, so now you have horse 1A, 1B, 1C, 1D, 1E, 2A, 2B, and so on. In each group, the fastest one is A and the slowest is E. Clearly, all horses that are D or E have no chance of being in the top three fastest, so you can ignore them. Now, race 1A, 2A, 3A, 4A, and 5A. For simplicity's sake, let's say that they finished in numerical order. That means that 4A and 5A are definitely not the top three fastest, and therefore, neither are 4B, 4C, 5B, or 5C. Ignore all of these. You also know that 1A is the fastest of all 25 horses. 1A beat 2A, who in turn beat 2B. 2C is therefore not in the top 3. 2A also beat 3A, so 3B and 3C are not in the top 3. This leaves you with five horses: 1B, 1C, 2A, 2B, and 3A. Race these five, and take the top two finishers. That is seven races
The Cruel King + Show Spoiler +There is a kingdom with population 51. There are 50 people, and the king. The king decides one day to line the fifty people up, and put either a red or blue hat on each person. The people are all facing the same way, so that the 50th person can see the hats of the 49 in front of him, but not his own. The first person, of course, can not see any hats. The king orders them to go down the line, from the 50th to the 1st, and guess the color of their own hat. If they get it wrong, the king will shoot them on the spot. If they get it right, they are allowed to live. Before they are lined up, they are allowed to devise a strategy to save as many lives as possible. What strategy do they use, and how many lives can they save? The strategy has to guarantee a certain number of people saved, so even though you could possibly save all fifty people if they all guess red, the worst case scenario is that they all die. Answer: + Show Spoiler +They can guarantee that 49 people are saved, and the other person has a 50/50 chance of living. Their strategy is for the 50th person (i.e. the first to guess) to look in front of him, and count the number of red hats. If it is even, he will guess red. If it is odd, he will guess blue. His guess has absolutely nothing to do with the color of his own hat. Let's say he guesses red. Then the 49th person knows that between him and the 48 people in front of him, there are an even number of red hats. He now counts the number of red hats in front of him. If it is odd, then the last red hat must be on his head. If it is even, then he is wearing a blue hat. He guesses correctly, and it just continues down the line. 50th guy is kind of screwed, and everyone in front of him lives.
The Cruel King, part 2 + Show Spoiler +The Cruel King is pissed that 49 (possibly 50?) people survived his first little activity. He devises a new one. He takes a group of seven people, and also seven index cards. On these index cards, he writes an integer between 1 and 7, inclusive, and he may repeat numbers. These seven people will be put into a room, and the card will be placed on their forehead, such that each person can see the other six cards, but not his own. Once in the room, they are not allowed to discuss anything. Then, all at once, they must each shout out a number. If at least one person guesses his own number, they are all saved. If nobody guesses their own number correctly, they are all killed. They are again allowed to devise a strategy prior to this challenge. What strategy do they use? Answer: + Show Spoiler +Each of the people will go into the room assuming that the grand total is a different least residue mod 7. There are exactly 7 least residues mod 7 (0 through 6), and of course, the total sum must be one of these. Then each person adds up the other six numbers, and determines what their own number must be given their respective assumptions. The person who has the right guess mod 7 will guess his own number right.
Oh, and this: http://www.cut-the-knot.org/Probability/LightBulbs.shtml
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On January 25 2011 17:18 Mayfly wrote:Correct, dear sir! (edit: for the tigers). The stones still need some work but you're very much on the right track. (Your proof is actually quite valid, but can be made even more general). edit 2: Some hints: The game is won by player number + Show Spoiler +. You need to already know that this game (and every game that has no ties) has a means to win it by perfect play, since it has no ties. It's not that proof I'm looking for here. Ah, I see. Hint (for those that want one): + Show Spoiler +The effect on the board of one player choosing a stone and the next player choosing a stone that is neither above nor to the right of the first is the same as if the first player had selected that second stone.
Last night, I didn't see how I could use that to prove that one player could force a win, but this morning I found the missing piece.
The winning strategy for player (named in spoiler above) falls into two parts: 1. + Show Spoiler +Select the top right stone 2. If this is part of a winning strategy, then this player has a win. If it is not, then + Show Spoiler +after player 1 selects the top right stone, player 2 must have a win (since there are no ties and it is not a winning strategy, it is a losing strategy). Then Player 2's next move is a winning move. The winning strategy for player 1 is simply to play that move first. This move will always be playable, as per the insight in the spoilered "hint".
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That was the solution I was looking for. Good job.
Exactly the same thing but another way to put it: + Show Spoiler +Imagine you're player 1 and that you're choosing the top-right-most stone. It is either a) the first step to a forced win, or b) a losing strategy. If the former then we have no problem. If the latter, then look at player 2's options. Can he make a move that you couldn't have made? Since you took the top-right-most stone, ANY stone he will pick would've also removed that stone had it been there. If indeed it is a losing strategy to pick the top-right-most stone, then you're to blame since you could have made the winning move player 2 will respond with. Hence, player 1 has a win here.
The winning starting move if you're curious is + Show Spoiler +to remove all but one stone in either the row or the column if the NxM board is asymmetric. If it is symmetric, just pick the stone top-right of the bottom-left-most stone.
I still don't know what this game is called, if anyone knows please tell me.
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On January 26 2011 01:03 Mayfly wrote:That was the solution I was looking for. Good job. Exactly the same thing but another way to put it: + Show Spoiler +Imagine you're player 1 and that you're choosing the top-right-most stone. It is either a) the first step to a forced win, or b) a losing strategy. If the former then we have no problem. If the latter, then look at player 2's options. Can he make a move that you couldn't have made? Since you took the top-right-most stone, ANY stone he will pick would've also removed that stone had it been there. If indeed it is a losing strategy to pick the top-right-most stone, then you're to blame since you could have made the winning move player 2 will respond with. Hence, player 1 has a win here. The winning starting move if you're curious is + Show Spoiler +to remove all but one stone in either the row or the column if the NxM board is asymmetric. If it is symmetric, just pick the stone top-right of the bottom-left-most stone. I still don't know what this game is called, if anyone knows please tell me.
Player 1 loses if there is only 1 stone . This is the only case where the induction of taking the 2nd player's move fails.
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The answer for the Forcefield Detainment one is so stupid, why doesn't the guard just say that he'll shoot the first one to leave?
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I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> )
The Pill Problem + Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do?
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On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? Hmm, if he chooses at random he has a 2/3 chance of guessing right, but that's not an answer. If he has enough time, he can + Show Spoiler +crush the pills to powder and mix them homogeneously so he knows exactly what he has (3 pills worth of 1/3 A 2/3 B) but your statement of the problem seems to exclude that. I can't think of any way to distinguish "indistinguishable" pills.
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On January 26 2011 04:37 youngminii wrote: The answer for the Forcefield Detainment one is so stupid, why doesn't the guard just say that he'll shoot the first one to leave? Because then they could all leave at once.
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On January 26 2011 06:47 qrs wrote:Show nested quote +On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? Hmm, if he chooses at random he has a 2/3 chance of guessing right, but that's not an answer. If he has enough time, he can + Show Spoiler +crush the pills to powder and mix them homogeneously so he knows exactly what he has (3 pills worth of 1/3 A 2/3 B) but your statement of the problem seems to exclude that. I can't think of any way to distinguish "indistinguishable" pills.
You're on the right track. Perhaps I made it sound like he doesn't have time to do anything; he does, just very limited time.
+ Show Spoiler +I retract that statement, you've basically solved it.
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On January 26 2011 06:47 qrs wrote:Show nested quote +On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? Hmm, if he chooses at random he has a 2/3 chance of guessing right, but that's not an answer. If he has enough time, he can + Show Spoiler +crush the pills to powder and mix them homogeneously so he knows exactly what he has (3 pills worth of 1/3 A 2/3 B) but your statement of the problem seems to exclude that. I can't think of any way to distinguish "indistinguishable" pills.
The Pill problem + Show Spoiler + That depends on the right now. If right now, then you just take two of the three for a 2 in 3 chance of survival.
Otherwise count out how many is left from either bottle, add the extra to the 3 you already have. The grind them and take half of the powder.
The timeframe seems tricky here..
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On January 26 2011 07:07 Luhh wrote:Show nested quote +On January 26 2011 06:47 qrs wrote:On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? Hmm, if he chooses at random he has a 2/3 chance of guessing right, but that's not an answer. If he has enough time, he can + Show Spoiler +crush the pills to powder and mix them homogeneously so he knows exactly what he has (3 pills worth of 1/3 A 2/3 B) but your statement of the problem seems to exclude that. I can't think of any way to distinguish "indistinguishable" pills. The Pill problem + Show Spoiler + That depends on the right now. If right now, then you just take two of the three for a 2 in 3 chance of survival.
Otherwise count out how many is left from either bottle, add the extra to the 3 you already have. The grind them and take half of the powder.
The timeframe seems tricky here..
Sorry for the ambiguity. He has time, just not enough to count all the pills in the bottle, but you're very close.
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On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do?
+ Show Spoiler +He counts the remaining pills in the bottles. He takes the one pill that's too much in one of the bottles. He halfs all four pills he has in his hands now and consumes a half of every pill, the other halfes the next time. Yay, the life is saved.
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On January 26 2011 07:17 Dandel Ion wrote:Show nested quote +On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? + Show Spoiler +He counts the remaining pills in the bottles. He takes the one pill that's too much in one of the bottles. He halfs all four pills he has in his hands now and consumes a half of every pill, the other halfes the next time. Yay, the life is saved.
+ Show Spoiler +Oh drat, I forgot to mention that he knows that he has one of A and two of B. Just saves him time, but you've solved it. Congrats :D
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On January 26 2011 07:17 Dandel Ion wrote:Show nested quote +On January 26 2011 06:37 thai_quan_doh wrote:I don't know what it's called, but it's probably my favorite riddle. (to anyone that knows how this goes, I apologize for butchering how it's told >.> ) The Pill Problem+ Show Spoiler +A man has been diagnosed with a terminal illness that can only be treated with two specific pills, we'll call them pill A and pill B. He must take one of each simultaneously at the same time of every day or else he will die. Since he cannot miss a day, he is only given a specific amount of pills which will last him exactly until he is given more pills; there are no leftovers or extras ever. These two pills are exactly the same in size, shape, color, taste, smell and weight; in other words, they are completely similar in appearance and are otherwise indistinguishable to all the senses. To alleviate this problem, the two bottles that contain pills A and B are sufficiently labeled to avoid confusion.
He takes the pills every morning at the same time. One morning, he wakes up late and has to rush to the bathroom so that he can take his pills and not die. In his haste, he mistakenly grabs one of pill A and two of pill B. He does not notice this until he is about to take the pills. He doesn't remember where each sat in his hand when he grabbed the pills. Taking all three right now would only prolong the inevitable; he would only die at a later date. He has to determine which is pill A and which is pill B right now, take them, then return the pill B to the bottle, or else he dies. What does he do? + Show Spoiler +He counts the remaining pills in the bottles. He takes the one pill that's too much in one of the bottles. He halfs all four pills he has in his hands now and consumes a half of every pill, the other halfes the next time. Yay, the life is saved. Ah, that's rather more elegant than my version of the answer, although the idea is similar.
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Red Eyes and Blue Eyes [tag: logic] + Show Spoiler + There is an island of monks where everyone has either brown eyes or red eyes. Monks who have red eyes are cursed, and are supposed to commit suicide at midnight. However, no one ever talks about what color eyes they have, because the monks have a vow of silence. Also, there are no reflective surfaces on the whole island. Thus, no one knows their own eye color; they can only see the eye colors of other people, and not say anything about them. Life goes on, with brown-eyed monks and red-eyed monks living happily together in peace, and no one ever committing suicide. Then one day a tourist visits the island monastery, and, not knowing that he's not supposed to talk about eyes, he states the observation "At least one of you has red eyes." Having acquired this new information, something dramatic happens among the monks. What happens?
+ Show Spoiler +Techniqually the answer to this is that 1 monk will commit suicide and no others, unless the tourist comes before the amount of days there are monks with red eyes.
The knowledge that would happen with n monks is that on nth day they will all commit suicide. They do not require the tourist to tell them this, thus only 1 monk is going to commit suicide because no one else has red eyes to prove that there would be doubt about whether they have them or not (if there is 1 he believes no one has red eyes because everyone else has brown).
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