Puzzles

[arkana]

From: http://3quarksdaily.blogs.com/3quark...lenge_you.html

Post the answer to ONLY ONE of the riddles so everyone gets a chance at figuring one out (ANSWERS IN WHITE PLEASE):

1. You are given two ropes and a lighter. This is the only equipment you can use. You are told that each of the two ropes has the following property: if you light one end of the rope, it will take exactly one hour to burn all the way to the other end. But it doesn't have to burn at a uniform rate. In other words, half the rope may burn in the first five minutes, and then the other half would take 55 minutes. The rate at which the two ropes burn is not necessarily the same, so the second rope will also take an hour to burn from one end to the other, but may do it at some varying rate, which is not necessarily the same as the one for the first rope. Now you are asked to measure a period of 45 minutes. How will you do it?

2. You have 50 quarters on the table in front of you. You are blindfolded and cannot discern whether a coin is heads up or tails up by feeling it. You are told that x coins are heads up, where 0 < x < 50. You are asked to separate the coins into two piles in such a way that the number of heads up coins in both piles is the same at the end. You may flip any coin over as many times as you like. How will you do it?

3. A farmer is returning from town with a dog, a chicken and some corn. He arrives at a river that he must cross, but all that is available to him is a small raft large enough to hold him and one of his three possessions. He may not leave the dog alone with the chicken, for the dog will eat it. Furthermore, he may not leave the chicken alone with the corn, for the chicken will eat it. How can he bring everything across the river safely?

4. You have four chains. Each chain has three links in it. Although it is difficult to cut the links, you wish to make a loop with all 12 links. What is the fewest number of cuts you must make to accomplish this task?

5. Walking down the street one day, I met a woman strolling with her daughter. “What a lovely child,” I remarked. “In fact, I have two children,” she replied. What is the probability that both of her children are girls? Be warned: this question is not as trivial as it may look.

6. Before you lie three closed boxes. They are labeled “Blue Jellybeans”, “Red Jellybeans” and “Blue & Red Jellybeans.” In fact, all the boxes are filled with jellybeans. One with just blue, one with just red and one with both blue and red. However, all the boxes are incorrectly labeled. You may reach into one box and pull out only one jellybean. Which box should you select from to correctly label the boxes?

7. A glass of water with a single ice cube sits on a table. When the ice has completely melted, will the level of the water have increased, decreased or remain unchanged?

8. You are given eight coins and told that one of them is counterfeit. The counterfeit one is slightly heavier than the other seven. Otherwise, the coins look identical. Using a simple balance scale, can you determine which coin is counterfeit using the scale only twice?

9. There are two gallon containers. One is filled with water and the other is filled with wine. Three ounces of the wine are poured into the water container. Then, three ounces from the water container are poured into the wine. Now that each container has a gallon of liquid, which is greater: the amount of water in the wine container or the amount of wine in the water container?

10. Late one evening, four hikers find themselves at a rope bridge spanning a wide river. The bridge is not very secure and can hold only two people at a time. Since it is quite dark, a flashlight is needed to cross the bridge and only one hiker had brought his. One of the hikers can cross the bridge in one minute, another in two minutes, another in five minutes and the fourth in ten minutes. When two people cross, they can only walk as fast as the slower of the two hikers. How can they all cross the bridge in 17 minutes? No, they cannot throw the flashlight across the river.

11. Other than the North Pole, where on this planet is it possible to walk one mile due south, one mile due east and one mile due north and end up exactly where you began?

12. I was visiting a friend one evening and remembered that he had three daughters. I asked him how old they were. “The product of their ages is 72,” he answered. Quizzically, I asked, “Is there anything else you can tell me?” “Yes,” he replied, “the sum of their ages is equal to the number of my house.” I stepped outside to see what the house number was. Upon returning inside, I said to my host, “I’m sorry, but I still can’t figure out their ages.” He responded apologetically, ‘I’m sorry. I forgot to mention that my oldest daughter likes strawberry shortcake.” With this information, I was able to determine all of their ages. How old is each daughter? I assure you that there is enough information to solve the puzzle.

13. The surface of a distant planet is covered with water except for one small island on the planet’s equator. On this island is an airport with a fleet of identical planes. One pilot has a mission to fly around the planet along its equator and return to the island. The problem is that each plane only has enough fuel to fly a plane half way around the planet. Fortunately, each plane can be refueled by any other plane midair. Assuming that refuelings can happen instantaneously and all the planes fly at the same speed, what is the fewest number of planes needed for this mission?

14. You find yourself in a room with three light switches. In a room upstairs stands a single lamp with a single light bulb on a table. One of the switches controls that lamp, whereas the other two switches do nothing at all. It is your task to determine which of the three switches controls the light upstairs. The catch: once you go upstairs to the room with the lamp, you may not return to the room with the switches. There is no way to see if the lamp is lit without entering the room upstairs. How do you do it?

15. There are two gallon containers. One is filled with water and the other is filled with wine. Three ounces of the wine are poured into the water container. Then, three ounces from the water container are poured into the wine. Now that each container has a gallon of liquid, which is greater: the amount of water in the wine container or the amount of wine in the water container?

[arkana]

Answer to #8 in white:

First weigh 6 coins (three each side), if the scale doesnt tip then weigh the remainding 2 coins (1 each side) to find the heavier one.

If the scale tips then we know the counterfeit coin is one of the three coins that were the heaviest. Put one of these three coins aside and put the other two coins on the scale (1 each side). If the scale tips then the heavier one is the counterfeit one, otherwise it is the coin you put aside.

[Da GOAT]

Answer to #8 in white:

First weigh 6 coins (three each side), if the scale doesnt tip then weigh the remainding 2 coins (1 each side) to find the heavier one.

If the scale tips then we know the counterfeit coin is one of the three coins that were the heaviest. Put one of these three coins aside and put the other two coins on the scale (1 each side). If the scale tips then the heavier one is the counterfeit one, otherwise it is the coin you put aside.

i got it another way. tho this is more exact

[jyms]

Wow what a great post, I love these.

1) use a watch, I couldn't think of anything else.
2) don't know (great start)
3) take the chicken across, go get the dog, take the dog across, bring back the chicken, take the corn over, go get the chicken. ( I had solved this one long ago)
4) 3
5) Obvious that it's 50% that the other is a girl, BUT the probability is only 33% that she has 2 girls.
6) take one from the mixed label. You know it's wrong, so it'll show you the colour. Place the label on the right box. Remove the wrong label and place on the third box. Remove worng label again and should match the jelly bean you removed from the first box. This was a great one.
7) increases
8) take out 2, weigh three vs three, if they weigh different, again weigh the heavy side, any two, if niether is heavy the third is the coin. that should be enough info
9) wine in the water is greater.
10) WOW, still working on it.
11) hhhmmmm
12) 3, 4 and 6
13) assuming you cant just let the plane crash and refuel at the halfway mark. You send two planes 1/4 around and refuel, while the first plane returns, the pilot flys around 1/2 way to the 3/4 mark. He meets up with another plane andrefuels and they fly back the last 1/4.
14) this took me the longest so far but I figured it out while at work. turn on the first switch for several minutes, then turn it off. then turn on a second switch, then head upstairs. if the light is on, bingo, if it's off, check to see if it's hot. This made my day figuring this out.
15) is the same as 9

I will keep tryuing on the others.

[thenonsequitur]

Okay, I figured out #11:

"Other than the North Pole, where on this planet is it possible to walk one mile due south, one mile due east and one mile due north and end up exactly where you began?"

There are lots of possible answers (an unlimited number actually).

* Consider the line of latitude near the south pole where the circumference of the earth is exactly one mile. You start at any point that is exactly one mile due north any point on this line of latitude. So, following the instrucitons: (i) you walk one mile due south, and end up on this line at a point we'll call p; (ii) you walk one mile due east, and since the circumference of the earth at this latitude is exactly one mile, you end up at p again; (iii) you walk one mile due north and end up exactly where you started.

* There is a line of latitude near the south pole where the circumference of the earth is exactly half of one mile. Again, start at one point due north of this line. Walk one mile south, and up at a point on the line we'll call p. Now, in walking east one mile you pass point p at half a mile, and end up on point p at the completion of the mile. Walk one mile north and end up where you started.

* Continuing this process, you can start at any point one mile due north of any point on a line of latitude near the south pole where the circumference of the earth is exactly 1/n miles, where n is a positive integer.

[thenonsequitur]

12) 3, 4 and 6

I don't think this is right. How did you arrive at this answer? The answer I got, and my reasoning, is below:

We know the following:

1. There are exactly three daughters.
2. The product of their ages is 72.
3. The sum of their ages is equal to a number that didn't disambiguate them (i.e. the man went outside, saw the sum, and still didn't know the answer).
4. The fact "the oldest daughter" likes strawberry shortcake disambiguated the results.

So, start by narrowing down the possible ages by premise (2). The prime factors of 72 are 2*2*2*3*3. By making all possible combinations of factors that result in three numbers, you get the following possible ages, along with the sum of their ages to the right.

2 2 18 = 22
2 3 12 = 17
2 4 9 = 15
2 6 6 = 14
3 3 8 = 14
3 4 6 = 13

By premise (3) we know that the sum alone doesn't disambiguate the result. And there are only two possible age sets that have the same sum (14). So the ages must either be "2, 6, 6" or "3, 3, 8".

But we know by premise (4) that there is a single oldest daughter. But the first age set has two oldest daughters.

So the ages of the girls must be 3, 3, and 8.

[thenonsequitur]

Question about #1: Are you allowed to use the sun? (if so it may be possible to use the one-hour measure of burning one rope, as well as some math, to consruct a sundial or something like a sundial out of the second rope, which can then be used to measure 45 minutes).

[jyms]

I just found the three numbers that would make the product of their ages 72. Since there is no mention of the same age or the number to use for the sum, I just used the only other clue their was, strawberry shortcake, a childrens cartoon. so the third child was not 18 if the three ages were 18, 4 and 1. so the oldest must still be a child leaving those numbers. But looking at your reasoning about the sum, you had one more clue I couldn't figure out. You sir are correct.

As for #1,you could start by burning the 1st rope at both ends this would only last one half hour and the start burning the second rope from both ends and the middle, there by separating the rope into 4 making it last a total of 15 minutes.

[thenonsequitur]

Hmm... Your mention of "1 4 18" made me realize I didn't even consider the factor of "1" as possible, which makes a number of more age groups possible. Namely: 1 4 18 = 23; 1 2 36 = 39; 1 3 24 = 28; 1 6 12 = 19; and 1 8 9 = 18 (and I guess, since it's the daughters' father, and the mother might be different, 1 1 72 is also a possiblity). But it looks all these sums are all singular, so it doesn't break the solution I came up with.

[AHiltz]

#5 25% since we have no idea wether or not the child seen with her is in fact hers

[jyms]

#5 25% since we have no idea wether or not the child seen with her is in fact hers

5. Walking down the street one day, I met a woman strolling with her daughter.

But your answer is still right. God damn, #2 and #10 are killing me. I'm not giving up.

[arkana]

Answer to #1:

Light the first rope at both ends and the second rope just at one end, when the first rope is burned out (30mins) light the other end of the second rope as well. Since there should have been 30 mins of burning left in the second rope, lighting the other end should cause it to burn out in 15 mins.

*Lighting a rope in the middle is not an option because one of the two ends could burn faster so it could catch up to the flame in the middle before the other one.

[thenonsequitur]

Okay, I figured out #2:

It's a deceivingly simple answer, but I think it's correct. Take any x coins from the pile of 50 and put them into pile A. Put all the other coins into pile B. Flip over each coin in pile A. Now you have two piles where the number of heads-up coins is the same.

I tried it for a number of cases, and it seemed to work for all of them.

[jyms]

You corrected my mistake on #1. The problem is I thought I solved it, now I got told the answer, that sucks. Good one though. One for you. How am i doing on the rest?

I think your off on #2, you are told for arguments sake you are told 8 are HU. If you have 2 piles say they are random. 1 has 10 coins with all 8 of the heads up in it and you don't know . No matter what you do to the other pile of 40 you cant match it without knowing the first pile has all 8.

[thenonsequitur]

I think your off on #2, you are told for arguments sake you are told 8 are HU. If you have 2 piles say they are random. 1 has 10 coins with all 8 of the heads up in it and you don't know . No matter what you do to the other pile of 40 you cant match it without knowing the first pile has all 8.

My solution does work, I think you are just misunderstanding what I said.

When I said "make a pile A of x coins" I meant to use the same x you are told. So, say that, as in your example, x=8. You start with 50 coins and are told 8 are HU. Now, take exactly 8 coins (any 8) and put them in another pile, and flip all 8 coins.

Now, consider all the possible cases:

1. All 8 HU coins were the 8 you took, leaving 0 in the other pile. You flip the 8 coins, and now have 0 HU in each pile. The same. Check.

2. 7 of the 8 coins you took are heads and 1 tails, leaving 1 heads in the other pile. Flip the 8 coins, and now 7 are tails, and 1 is heads. You now have 1 HU coin in each pile. Check.

3. 6 of the 8 coins you took are heads and 2 tails, leaving 2 heads in the other pile. FLip the 8 couns, and now 6 are tails, and 2 are heads. You now have 2 HU coins in each pile. Check.

Etc... The process can be continued for the other 6 cases. In each case, you get the same number of heads.

Also, it seems to work for every x, not just x=8.


A more general formulation follows:

You have x coins that are HU. Split all the coins into two separate piles, A, and B. Let the size of pile A be sa, and the size of pile B be sb. There will always be a number of HU coins in A, call it na, and a number if B, call it nb. Note that x = na + nb. So nb = x - na, and na = x - nb. Now if you flip all the coins in pile A, all that were tails are now heads and vice versa, so you now have (sa - na) heads up coins in pile A (and still have x-na H coins in pile B). So if you want to make the number of HU coins equal, make these items equal: sa - na = x - na. Solve and you get sa=x. So set the size of pile A equal to x, and you will get the same number of HU coins for each pile, regardless of x, and regardless of the number of HU coins that end up in each pile before the flipping.

[jyms]

I think your off on #2, you are told for arguments sake you are told 8 are HU. If you have 2 piles say they are random. 1 has 10 coins with all 8 of the heads up in it and you don't know . No matter what you do to the other pile of 40 you cant match it without knowing the first pile has all 8.

My solution does work, I think you are just misunderstanding what I said.

When I said "make a pile A of x coins" I meant to use the same x you are told.
.

you failed to mention this and that is waht i was working on. You beat me

[Aces]

Answer #10

(10) and (1) cross the bridge = 10min
(1) goes halway to meet(5), they continue to cross = 5min
(1) goes halfway to meet(2), they continue to cross = 2min

Total = 17min

Maybe?

[Setzy]

Not sure if I'm circumventing a bigger idea, but..

#9. Wine is comprised of more than 50% water, so there is more water in the wine than wine in the water.

#10. Wait till daylight. Ten minute and five minute hiker cross together, then two minute and one minute hiker cross together.

[sandstorm]

Hmm, another answer for #2:
I didn't really get that you KNEW how much x is. Oh well, but I thought hey, why not take as many coins as you like and put them in two piles, then put them on the table on the sides, so NO coins are heads up. Easy, eh?

1&2 down

[sandstorm]

14. You find yourself in a room with three light switches. In a room upstairs stands a single lamp with a single light bulb on a table. One of the switches controls that lamp, whereas the other two switches do nothing at all. It is your task to determine which of the three switches controls the light upstairs. The catch: once you go upstairs to the room with the lamp, you may not return to the room with the switches. There is no way to see if the lamp is lit without entering the room upstairs. How do you do it?

Flip switch 1 and leave it on for a while
Flip back switch 1 and then switch number 2, then leave

If the bulb is lit, it's nr 2
If the bulb is off but warm it's nr 1 (hee hee)
If the bulb is just off, it's 3.

[nickthefool]

3.
Takes chicken across
Comes back alone
Takes dog across
Brings chicken back
Takes corn across
Comes back alone
Takes chicken across

8.
1st use of scale - put 3 coins on each side.
If they are level,
2nd use of scale - compare the 2 coins not weighed in the first use, the heavier 1 is counterfeit.
If they are not level,
2nd use of scale - take coins from the 3 that were heavier in the 1st weighing, and compare them. If they are equal, the other one is counterfeit. If not, the heavier one is.

[jyms]

So except for #10 the hikers, we have them all?

[sandstorm]

Someone solve number 5 and explain plz

Is it the daughter he's talking to? How do we know that in that case?

[jyms]

Someone solve number 5 and explain plz

Is it the daughter he's talking to? How do we know that in that case?

I thought I did.

5)Obvious that it's 50% that the other is a girl, BUT the probability is only 33% that she has 2 girls.

The probability is only 1 in 4 that she has two daughters. 2 boys, a boy and a girl or two girls. The odds of her other being a girl is 50% since we know that one is already a girl. Probability is the chance "before" any facts are known.

[thenonsequitur]

I don't agree with either of the #9 answers presented so far. I figure it like below:

"9. There are two gallon containers. One is filled with water and the other is filled with wine. Three ounces of the wine are poured into the water container. Then, three ounces from the water container are poured into the wine. Now that each container has a gallon of liquid, which is greater: the amount of water in the wine container or the amount of wine in the water container?"

Answer: The amount of wine in the water container is equivalent to the amount of water in the wine container.

Note that 1 gallon = 128 ounces.

So first three ounces of wine are poured into the water. So the wine container now contains 125 ounces of wine, and the water container now contains 128 ounces of water and 3 ounces of wine.

This is a new mixture. Now we want 3 ounces of this new mixture. Note that there are a total of 131 ounces in the counter, so 3/131 is the ratio we want to take away. To figoure out what comprises three ounces of the mixture, just take this ratio of each substance. So 3 ounces of the mixture is 3/131*128 = 384/131 ounces of water and 3/131*3=9/131 ounces of wine (note that the numerators, 384 and 9, add up to 3 times 131, so it does add up to 3 ounces).

So after removing this amount, the water container now contains 128-(384/131) = (16768/131)-(384/131) = 16384/131 ounces of water and 3-(9/131) = (393/131)-(9/131) = 384/131 ounces of wine.

Now we add the 3 ounces of mixture back to the original wine container, which previously had 125 ounces of wine. Note that this is 16375/131 ounces of wine. After adding the mixture, the wine container now contains 16375/131+9/131 = 16384/131 ounces of wine and 384/131 ounces of water.

So it works out that there is exactly as much wine in the water container as water in the wine container (the exact amount that was exchanged is 384/131).

[jyms]

Nope. To simplifie it without all the fractions and to appease the canadians, a 1 gallon container has 3785 milliliters.
so 3 ounces is 88.5 mils
If you put 88.5 (3 oz)into 3785 mils. you have 88.5/3873.5 in the water container
when you remove 88.5 mils. of that fluid you have .0228 mils of wine in the 3 oz. of water.
leaving 88.4772 wine + 3696.5228 =3785mil.s (1 gal) water in the gallon water container
When you put the 88.5 (3 os)mixture of .0228 wine and 88.4772 water in the wine container it leaves 3785 (1 gal.) - 88.5 (3 oz)=3696.5 wine + .0228 wine for a total of 3696.5228 of wine and 88.4772 water and the water jug with 88.4772 wine in it And if I had done this from the beginning I'd of ben as right as you.

[arkana]

#5 25% since we have no idea wether or not the child seen with her is in fact hers

Wrong, the question states you meet a woman that is walking with "her daughter". The answer is actually 50% because of the way the problem was stated (a bit ambiguous). If the question was: "a woman has two daughters and when asked if at least one of them is a girl she answered yes. what is the probability she has two girls?" then it would be 1/3.

[arkana]

Answer to #10:

1&2 cross in 2 minutes. 1 returns in 1 minute. 5&10 cross in 10 minutes. 2 returns in 2 minutes. 1&2 cross in 2 minutes. Total: 17 minutes.

[jyms]

Answer to #10:

1&2 cross in 2 minutes. 1 returns in 1 minute. 5&10 cross in 10 minutes. 2 returns in 2 minutes. 1&2 cross in 2 minutes. Total: 17 minutes.

Great job. Are you figuring these out or are the answers posted somewhere. LOL

[Setzy]

I knew my answers were too realist. Nice work.

[thenonsequitur]

Answer to #10:

1&2 cross in 2 minutes. 1 returns in 1 minute. 5&10 cross in 10 minutes. 2 returns in 2 minutes. 1&2 cross in 2 minutes. Total: 17 minutes.

Nice! This one has been bugging me. Good job.

[bigred]

42

[arkana]

Answer to #10:

1&2 cross in 2 minutes. 1 returns in 1 minute. 5&10 cross in 10 minutes. 2 returns in 2 minutes. 1&2 cross in 2 minutes. Total: 17 minutes.

Great job. Are you figuring these out or are the answers posted somewhere. LOL

I figured out my answers but they have since been posted. You can find them by following the link at the top of my post.

[arkana]

Okay, I figured out #2:

It's a deceivingly simple answer, but I think it's correct. Take any x coins from the pile of 50 and put them into pile A. Put all the other coins into pile B. Flip over each coin in pile A. Now you have two piles where the number of heads-up coins is the same.

I tried it for a number of cases, and it seemed to work for all of them.

Very nice