EPT
Brainteasers/Math problems - Page 3
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mcc
Czech Republic4646 Posts
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thoradycus
Malaysia3262 Posts
On April 10 2011 00:24 Wesso wrote: + Show Spoiler + That's assuming all the circles have no overlap which is impossible + Show Spoiler + Well the question states "in some way",so i guess we have to follow the assumption that it can somehow fill up the whole rectangle without overlaps.A somewhat unrealistic question if u ask me. | ||
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oxidized
United States324 Posts
4. A very accurate clock has an hour hand and a minute hand. Both hands are (infinitely) thin. At 12 noon, the two hands coincide exactly. What is the next (exact) time at which the two hands will again coincide? Obviously it will occur after 1:00. The minute hand will be slightly past 5 as someone mentioned earlier in thread. This is just a basic system of equations math problem. Variables: HA = hand angle, MA = Minute Angle, m = number of minutes past 1 HA = 30 + (30 m / 60), where m is the number of minutes that have passed. MA = 6 m. Condition: HA = MA HA = 30 + MA /12 => MA = 30 + MA / 12 Solving: 11 MA = 360 => MA = 360 / 11 m = MA / 6 => m = 60 / 11 Therefore, they overlap in 1 hour and 60/11 minutes (approx 5 min 27 sec) + Show Spoiler [5] + 5. (my favorite ) Suppose a rectangle can be (in some way) entirely covered by 25 disks, each of radius 1. Can the same rectangle be covered by 100 disks of radius 1/2? Prove your answer. Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap. + Show Spoiler [10] + On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)? Cool Question. Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them. The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition. If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see. If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color. The viziers will agree on all this on the night before the hat day. More elegant solution (same idea): see Murderotica's post a couple posts down. | ||
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Slunk
Germany768 Posts
On April 10 2011 01:21 Starfox wrote: + Show Spoiler + Nope, the guru plays a roll in the solution, it's a rather famous riddle, the numbers could be anything though, the logical conclusion goes for every number of people + Show Spoiler + But the guru only speaks once and tells something that everybody knows allready. | ||
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Murderotica
Vatican City State2594 Posts
On April 10 2011 01:27 mcc wrote: Just to add for Day9's strings aka problem 8. The standard solution provided still has one nitpicky assumption that speed of burning is independent of the direction of burning. I did not think about it too much, but I suspect that without that assumption it has no solution, but if someone has any, I would love to hear it, I will try to prove that there is none in the meantime There is a solution for the string to only burn in one direction: + Show Spoiler + 1. Put the two strings side by side. 2. Fold one string in half, folding to the side. 3. Fold that string in half again, folding over the top. 4. This will cause there to be 2 quarters of string side by side and 2 on top of those 2. Unfold the string from the quarter that is adjacent to the second string, and put the unfolded string perpendicularly over the other string. 5. Light the extended string. When the flame ignites the second string which is on top of the first (perpendicularly), the allotted time has passed. | ||
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stepover12
United States175 Posts
On April 10 2011 01:27 thoradycus wrote: + Show Spoiler + Well the question states "in some way",so i guess we have to follow the assumption that it can somehow fill up the whole rectangle without overlaps.A somewhat unrealistic question if u ask me. The original covering by 25 disks must have overlaps. And the covering by100 disks (is it exists) should have overlaps as well. | ||
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Murderotica
Vatican City State2594 Posts
On April 10 2011 01:31 oxidized wrote: + Show Spoiler [5] + Yes, the circles cannot overlap. The solution is to note that you can half each dimension of the original rectangle filled with cirlces, and the circles will have their radius cut to 1/2, with the original orientation preserved. Then construct four of these small rectangles and preserve the orientation of the 1/2 circles. Put the four rectangles together in such a way to make the original rectangle dimensions. Then you have a pattern of 1/2 size circles which will always properly fill the rectangle (4 times the original number of cirlces). No overlap as long as the original had no overlap. + Show Spoiler [10] + On April 10 2011 00:36 ILOVEKITTENS wrote: Sultan summons all of his viziers. He says "Tomorrow I am going to put all of you in a line and place a hat on each of your heads. The hat will either be red or blue. You will not be able to see the hat on your head. However, because you are my royal viziers, you must be able to tell me what color hat is on your head. Only one of you may be wrong - otherwise, you all die. You can tell me the color of your hat in any order, and you are only allowed to say the color and nothing else - no communication with other viziers." How do the viziers keep their jobs and their lives (what is their strategy)? Cool Question. Since one vizier can be wrong, the first vizier needs to say red or blue in a way that will give information to the rest of them. The first vizier needs to give information on the parity (odd/even) of the red or blue on the hats that he sees. That way, the rest of the viziers can count the hats, and determine his own hat color based on the parity condition. If there are an even number of viziers (say 100), the first vizier will see an odd number of total hats (say 99) - that means you have an odd number of one hat color, and an even number of the other hat color. The first vizier to go says the hat color of whichever he sees as odd. Then the rest of the viziers can deduce their own hat color based on what they see. If there are an odd number of viziers (say 99), the first vizier will see an even number of total hats (say 98). Even+even = even or odd+odd = even. Thus, the viziers agree on a convention - for example: say blue if it is odd plus odd, and say red if it is even+even. With this parity information, the rest of the viziers can determine their own hat color. The viziers will agree on all this on the night before the hat day. + Show Spoiler + + Show Spoiler + + Show Spoiler + 10. + Show Spoiler + That solution can be simplified. We do not need to know whether the amount of viziers is even or odd. They can assert that the number of blue hats will be odd, whether it is or not. So, if the first vizier sees an odd number of hats, he will say he has a red hat to keep the number odd. The next vizier will know that the guy behind him said that his hat was red because he sees an odd number of blue hats ahead of him, and from this information he can deduce what hat is on his head after viewing if the number of hats ahead of him is odd or even. This is more elegant but you were of course still correct. | ||
Kipsate
Netherlands45349 Posts
You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room? | ||
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Wr3k
Canada2533 Posts
+ Show Spoiler + For example, if we had 2 blue (A & B) 2 brown (C & D) 1 green (guru): Everyone knows: -at least 1 blue (Guru statement is key to this problem) -Therefore, we can assume that if there is a person who sees no others with blue eyes, he will leave knowing he has blue eyes. NIGHT 1: Nothing happens A & B: See 1 blue 2 brown 1 green. Know that there are at least 1 blue. No conclusions can be made. C & D: See 2 blue 1 brown 1 green. Know that there are at least 1 blue. No conclusion can be made. Guru:: 2 blue 2 brown. No conclusion. NIGHT 2: A & B leave. A & B: See 1 blue 2 brown 1 green. Noone left on the 1st night, therefore they can imply there are >1 persons with blue eyes & therefore they must have blue eyes. Both of them leave. C & D: See 2 blue 1 brown 1 green. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes. Guru: 2 brown 2 blue. Nobody leaving the island on the first night provides no useful information to them since they have already seen two individuals with blue eyes. NIGHT 3: C & D 1 green 1 brown. C & D can conclude that their eyes were not blue since A & B have left together. This doesn't help since both C & D cannot deduce if there are 2 or 1 brown eyed people remaining. For all they know their own eyes are red and as such the information from the 1st and 2nd nights is useless. Guru: The guru cannot get any further useful information either since there is no information pertaining to green eyes available to him. Given this example we can begin to see that if there are X blue eyed people they will leave the island on night X, but only if they know there is at least one blue eyed person. We can then conclude that all of the blue eyed people will leave on the 100th night. | ||
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Murderotica
Vatican City State2594 Posts
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it: You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room? Uh well... These switches are outside the room right? | ||
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oxidized
United States324 Posts
On April 10 2011 01:39 Murderotica wrote: 10. + Show Spoiler + That solution can be simplified. We do not need to know whether the amount of viziers is even or odd. They can assert that the number of blue hats will be odd, whether it is or not. So, if the first vizier sees an odd number of hats, he will say he has a red hat to keep the number odd. The next vizier will know that the guy behind him said that his hat was red because he sees an odd number of blue hats ahead of him, and from this information he can deduce what hat is on his head after viewing if the number of hats ahead of him is odd or even. This is more elegant but you were of course still correct. + Show Spoiler + Yep, good answer! EDIT: On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it: You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room? + Show Spoiler + I have seen this before and I think it is a terrible question. It is not a math/logic type brainteaser. The solution involves keeping one light on for a while and then turning it off so it remains hot when you go in, so you can feel it (and turn on another one before going in). Kinda lame imo | ||
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Dullahan
United States248 Posts
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it: You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room? + Show Spoiler + Don't you also need to specify that the light bulbs generate heat when lit? | ||
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Tunks
United Kingdom8 Posts
On April 10 2011 01:42 Kipsate wrote: I have 1, I found it to be a little farfetched when I knew the answer but it sticked with me due to that, I am interested to see if people can solve it: You have 3 switches, a room and a light. 1 of these buttons turns on the light in the room, you are only allowed to go in the room once. How do you know which button turns on the light in the room? + Show Spoiler + Turn the first switch on and wait for a few minutes, then turn it off again. Turn the second switch on and enter the room. If the light is on you know it's switch 2 If the light is off but the bulb is hot it's switch 1 If the light is off and the bulb is cold it's switch 3 | ||
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funk100
United Kingdom172 Posts
this there is this village beside a mountin where all justice is decided by the accused picking one ball out of a bag with two in - one white one black- they then show the ball to the audience and justice is administerd. if the criminal picks the black ball they are thrown off the cliff (where the trails take place) if they pick the white ball they can walk free. let us suppose that there is a man who has commited 10 murders but has continually, through chance, always picked the white ball and allways walked free. On the night of his 11th murder the brother of the murdered gets pissed as he wants justice, so he switches the bag so both the balls are black and so, he reasons he can get revenge. the murderer hears of this from the brother as he put his plan as his fb status and the murderer is a mutuall friend of him. On the day the murderer manages to walk free from the cliff after the village agree he picked a white ball, how? REMEMBER/rules 1) there are 2 black balls in the bag and they are not changed 2) he does not have a trick ball anywhere - he chooses a ball from the bag 3) the ball he chooses is the one he is represented by 4)where he is | ||
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Kazius
Israel1456 Posts
1 hour 34 minutes left to the TSL, let's see how my speed is (including reading the questions, writing down the answers, and this): 1. + Show Spoiler + Fill the 5ltr jug, use it to fill the 3ltr jug. You now have 2ltr left in the 5 ltr jug. Empty the 3ltr jug, pour the 2ltr from the 5 ltr jug into it. The 3ltr jug now has place for 1ltr. Fill the 5 ltr jug, and then, with it fill the 3ltr jug until it is full. You now have 4ltrs in the 5 ltr jug. 2. + Show Spoiler + It is possible. You take one coin from the first, 2 coins from the second, three coins from the third, etc until 10 coins from the 10th. If you weigh the coins you took, you will get a weight of 55gr (55 coins), -0.1gr for each counterfit coin. If you have one counterfit coin, it will weigh 54.9, for 2 it'll be 54.8, etc until 10 will be 54gr. According to the number of counterfit coins, you will know from which bag you took them. 3. + Show Spoiler + Flip the 7 and 11 minute hourglasses simultaneously. Start cooking once the 7min one runs out. That leaves 4 minutes until the 11 minute hourglass is empty. Once it empties 4 minutes later, flip it over. Once it is done, you have been cooking for exactly 4 + 11 = 15 minutes. 4. + Show Spoiler + Let's assume that they move at an exact uniform speed (won't skip eachother). In the next 12 hours, they will meet 11 times (once at 1pm and a bit, once at 2pm and a little more, etc, until the meet again exactly at 12am). Therefor, in 12 hours they meet 11 times. Since they move at uniform speeds, they must meet once every 12/11 hours, meaning 1 hour, 5 minutes, 27 seconds, and 3/11ths of a second (if my arithmetic is correct). 7. + Show Spoiler + It can be found in 2 comparisons like this: Split into 3 piles of 3, compare 2 of them on the scale. If one is lighter, then in contains the coin. Otherwise, it is in the 3rd pile. Hence, the coin is in the lightest pile. Take 2 coins from that pile, compare them on the scale, if one is lighter, that is the coin you are looking for, otherwise, it is the coin that wasn't on the scale. 8. + Show Spoiler + You take the first string and light it on both sides so it will burn twice as fast. You light the other string on one side at the same time. Once the string lighted on both sides finishes burning, exactly 30 minutes have passed, so start burning the other string on the other side, so it will burn twice as fast. Since the remaining 30 minutes of burn times have been halved, it will finish burning in another 15 minutes. Once it finishes burning, 45 minutes have passed. 9. + Show Spoiler + Let n be a prime number over 30. Let m be the remainder of dividing n by 30, and p is the result rounded down. m is therefor a non-prime number under 30. Let's assume that m is not a prime number. Therefor, n = 30*p + m. Hence, p = (n-m)/30. If m is even, therefor, n has to be even, since an odd number minus an even number is an odd number, and p is by definition a whole number. If m is divisible by 3, then so is n. If m is divisible by 5, so is n, for the same reason. Hence, m has to be a multiplication of prime numbers greater than 5. Hence, m is equal to or greater than 7*7 = 49, which is impossible as it is the remainder for division by 30. Hence, the remainder has to be a prime number. 5, 6 and 10 will take time I don't have if I want to get stuff done before TSL, so it will have to wait. Edit: dammit, just thought of this: 5. + Show Spoiler + Sure they can. The rectangle can be filled with 25, so a rectangle with half the height and half the width can be filled with circles with half the radius in the exact same way. Therefor, you can put 25 of each of the smaller discs in each quarter of the rectangle using the exact same strategy to fill it. Feel stupid that I didn't get that instantly :/ Edit 2: OK. The first 1-4 and 7-9 were all instant, and came to me after typing the rest. So that's 21 minutes. ARG!!@# 3 more now! Let's take another crack at this: 10. + Show Spoiler + The first vizier says his hat is red if he counts an odd number of blues on the others. Therefor, if he is correct, everyone knows there is an odd number of blues and can deduce that if he sees an even number of blues his hat is blue, and otherwise it is red. If he is incorrect, therefor there is an even number of blue hats, and the other viziers can (in similar fashion) deduce the color of their hat. 12. + Show Spoiler + You should switch. At your first choice, you had a 1 in 3 chance to get it right, and a 2 in 3 that is one of the others. Since he opened one of them, it is a 2 in 3 chance it is behind the other door 13. + Show Spoiler + Rephrasing the question: Can any natural number be uniquely expressed in base 2? Answer: yes, those are called binary numbers. 14. + Show Spoiler + The first line can't cross any. The second line can at most cross the first. The third can cross the other two at most. And so on. Hence: the maximum number of crossings is 0+1+2+3+...+9, therefor, 45 possible intersections. Hooray for another 10 minutes spent! 6 and 11 will take some time, and I want to get food ready for TSL :D | ||
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gyth
657 Posts
Pour 3L into the 5L, pour another 2L in leaving 1L in the 3L container, dump out the 5L, pour the 1L into the 5L, then another 3L for a total of 4L. 2.+ Show Spoiler + 1 coin from first bag, 2 from second, 3 from third... number of tenths of a gram you are off is the bag of counterfeit coins. 1 from first bag, 2 from second, 4 from third, and 8 from fourth. Convert tenth gram difference to binary and those are the fake bags. 3.+ Show Spoiler + Start both, flip the 7m when it runs out after 7m, flip it again in 4 minutes when the 11m runs out, it will then run out 4 minutes later for a total of 15 minutes. 4.+ Show Spoiler + In 12 hours this happens 11 times, so one hour, 5 minutes and 300/11 seconds, 1:05:27.2727... 5.+ Show Spoiler + 100 disks of radius 1 can cover 4 same sized rectangles, shrinking everything to .5 scale you have 1 rectangle of the original size covered by 100 disks of radius .5 6.+ Show Spoiler + On the first night if someone saw noone else with blue eyes they would know they had the blue eyes and they could leave. But since that doesn't happen everyone knows that noone sees 'no one with blue eyes'; ie. everyone sees at least 1 person with blue eyes. So on the second night someone who saw only one person with blue eyes would know that they also had blue eyes... So one the 100th night everyone who sees 99 people with blue eyes will know that they have blue eyes, so all 100 of them leave that night. 7.+ Show Spoiler + Weigh 3 against 3, if one weighs less it contains the counterfeit. If it balances then the counterfeit is in the 3 not weighed. Weigh 1 vs 1 of the last 3. 8.+ Show Spoiler + Light three of the ends at the same time. The one with 2 lit ends should burn in 30 minutes. Then light the other end of the remaining string and it should burn out in 15 minutes. 9.+ Show Spoiler + If the remainder was a multiple of 2,3,5 then it wouldn't have been prime before dividing by 30. So the only remainders possible when starting with a prime number are 7, 11, 13, 17, 19, 23, 29 which are all prime. 14.+ Show Spoiler + 1+2+3+4+5+6+7+8+9=45 15.+ Show Spoiler + If we group the 100 numbers by their mod 7. The difference between members in a group would be a multiple of 7. By pigeon hole, the smallest the largest group could be is 15 members. 23.+ Show Spoiler + 12 coins, 1 of which is either heavy or light EEEEEEEEEEEE (24 possibilities), measure 4v4 they balance, the 4 others are either heavy or light ___EEEE (8 possibilities), measure 3 versus 3 legit ___ they balance, the one you didn't measure is either heavy or light ______E (2 possibilities), measure it versus 1 legit ___ they don't, you have 3 heavy or 3 light ______hhh or lll (3 possibilities), compare 2 ______ they balance, the third is the fake ______ they don't, the one that matches heavy or light is the fake they don't, one of the 4 is heavy or one of the other four is light ___hhhhllll (8 possibilities), measure hhl versus hhl ___ they balance, one of the 2 lights you didn't measure is fake ______ll (2 possibilities), measure versus each other ___ they don't, one of the 2 heavies was heavy or the opposite light was light ______hhl (3 possibilities), measure the heavies versus each other ______ they balance, the light was light ______ they don't, the heavy one is heavy | ||
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NB
Netherlands12045 Posts
its not that simple to solve it since you will have to alternate the coins in the set you used to weight the 1st time. The REAL #7 should be that you DONT know if the fake coin weight MORE or LESS than the real coins. It makes the problem WAYYYYY more complicated but still solve-able (hoping the OP gona put this one up :D) There are 12 coins. One of them is fake: it is either lighter or heavier than a normal coin. Find the fake coin and tell if it is lighter or heavier by using a balance the minimum number of times possible. (answer is 3) | ||
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esla_sol
United States756 Posts
+ Show Spoiler + we know immediately that the hands will touch again sometime after one. so i set up two equations, y=x which is the movement of the minute hand at one, and y=x/12 + 5 which is the hour hand at one. from there i just found out where the two lines intersect, which happens 5.45 minutes after the clock hits one. so the answer is 65 minutes 27 seconds. | ||
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stepover12
United States175 Posts
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Tunks
United Kingdom8 Posts
You are in the final round of a game show and are shown 3 doors. You will win whatever is behind the door you eventually choose. Behind 1 door is a car, and behind the other 2 are goats. You make your original choice but don't open it straight away and the presenter opens one of the other 2 doors you didnt choose to reveal a goat. He then gives you the chance to switch to the other remaining closed door, or to open your original choice. Should you switch? For the OP and others, yes this is the monty hall problem. | ||
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