## 1. Stranded on an Island with a Dragon

This is probably my favorite puzzle of all-time, as it's very clever and different from all the usual puzzles -- you are stranded on an island with a dragon. This island has seven different wells at different elevations. Each of these wells is poisoned and if you drink from a well, you will die in half an hour, unless you drink from a higher well. In other words, if you drink water from Well #1, you can cure yourself by drinking water from Well #2, or any other higher well.

Unfortunately for you, the highest well is located at the top of a very high cliff, where only the dragon can reach it.

You and the dragon decide to have a duel -- each of you will bring a glass of water to the duel, which the other will have to drink. If you assume that the dragon is rational, what do you do to ensure that both A) you survive, and B) the dragon dies?

**First, consider how you will survive the duel. The dragon can reach Well #7, and he knows that you cannot cure yourself from that water, so you can assume that he will bring water from that well. The trick is to drink water from another well**

Second, consider how you will kill the dragon. The dragon knows that the water from Well #7 will cure any of the other wells, so you can assume that he will drink water from that well after the duel. The trick here is to bring him unpoisoned water (from the surrounding ocean). Thus, when he drinks the water from Well #7, he'll poison himself and be unable to cure it.

*before*the duel. Thus, when you drink the water from Well #7, that water will cure you!Second, consider how you will kill the dragon. The dragon knows that the water from Well #7 will cure any of the other wells, so you can assume that he will drink water from that well after the duel. The trick here is to bring him unpoisoned water (from the surrounding ocean). Thus, when he drinks the water from Well #7, he'll poison himself and be unable to cure it.

## 2. The Classic Hat/Prisoner Puzzle

Four prisoners are lined up in a row, with a wall separating the first three prisoners from the fourth. Each has a hat, two of which are black and two of which are white. Each of the prisoners know that there are exactly two white hats and exactly two black hats, but they are only able to see the hats in front of them, as seen in the picture. The prisoners will all be released if any one prisoner is able to correctly identify the hat he is wearing. How do the prisoners escape?

**First, consider prisoner A. He can see the hats of prisoners B and C. If those two prisoners are wearing the same color hat, then prisoner A knows he must be wearing the opposite color. Thus, he will immediately be able to identify his own hat color. In contrast, if B and C are wearing different colored hats, A will not be able to tell his own. However, prisoner B knows this, and thus, if prisoner A does not identify his own hat color, prisoner B knows he must have a different color hat than prisoner C, and will thus be able to identify his own hat color. Thus, either way, the prisoners will escape.**

## 3. Another Hat/Prisoner Puzzle

This is a very neat variant of the classic prisoners with hats puzzles: 100 prisoners will be lined up in a row, and each will have a black or white hat placed on their head. They will be asked one at a time, in line starting from the back, which color hat they are wearing. If they guess correctly, they will be released. Each prisoner is unable to see their own hat, but can see the hat of all the prisoners in front of them in line. Thus, the 100th prisoner knows the hat color of everyone else, etc. The prisoners are able to decide a plan in advance to maximize how many prisoners are saved. What tactic should they use, and how many people will be saved?

For example, one strategy would be for every other individual to say the hat color of the person in front of them, and for the following person to simply state that color. In this way, every other person would be saved, and the remaining individuals will be saved only if they happen to have the same hat color as the person in front of them. However, it is possible to do much better than this...

**This puzzle is very cute because it has a simple trick: the prisoners simply agree ahead of time that the last person in line will say black if there are an odd number of black hats in front of him, or white if there are an even number. The next person in line then simply counts the number of black hats in front of him, and then knows whether his must be black or white. For example, if the last person said black, and there are an even number of black hats in front of the next-to-last person, the next-to-last person's hat must be black, since the total must be an odd number. The next person can proceed in the same way, and every person will be able to deduce the color of their own hat. The only person not saved will be the first person, who must rely on simply being lucky.**

## 4. Taking Your Pills

Every day, you have to take two pills, Pill A and Pill B. It's vitally important that you take exactly one of each type at exactly the same time. One day, you pour one of the A pills into your hand, but accidentally pour two of the B pills into your hand. The pills are absolutely indistinguishable from one another in every way. What can you do to ensure that you still continue to take exactly one of Pill A and exactly one of Pill B every day?

**You currently have one A pill and one B pill in your hand. If you take another A pill, you have two A pills and two B pills. The trick is to now cut all four of the pills in half, keeping one half of each pill in one pile, and the other half in another pile. Now, each pile contains two halves of an A pill and two halves of a B pill. Thus, you'll still take one complete A pill every day and one complete B pill every day.**

## 5. Heads or Tails?

You are placed into a completely dark room with a pile of coins on the floor. You are told that 50 of these coins are heads and 50 of them are tails, but you are completely unable to see which are which. You are told to make two piles, such that the two piles have equal numbers of heads. How do you do this?

**You take 50 of the coins into one pile and 50 of the coins into the second pile. You then flip over all of the coins in the second pile. The two piles will now have the same number of heads. For example, if you happened to have picked 42 heads and 8 tails into the first pile, then the second pie has 8 heads and 42 tails. When you flip over the coins in the second pile, it now has 42 heads and 8 tails, just like the first pile.**

## 6. Find the Thief

A thief has stolen your treasure! The thief ran off into the woods, into a network of 17 caves all in a row, numbered #1-17. Every day, the thief will move to an adjacent cave. For example, if he starts in Cave #3, he will move to either Cave #2 or Cave #4 the next day. You can check one cave per day, but you have to get to your treasure within a month. Is this possible? How do you do it?

**The trick with this puzzle is that every day, the parity of the thief flips, i.e. if he is in an even-numbered cave on the first day, he is in an odd-numbered cave on the second day. Thus, you can start out by assuming he begins in an even-numbered cave.**

On the first day, you check Cave #2. If he is in Cave #2, you'll catch him on the first day. If he's not in Cave #2 (but he did start in an even-numbered cave), then on the next day he must have moved into an odd-numbered cave between #3-#17 (since the only way he could get to #1 is to start out in #2, which you checked). So, on the next day, you can try Cave #3. If he's not there, then on the next day he must be in an even-numbered cave between #4-#16. Thus, you check Cave #4. In this manner, you keep checking adjacent caves until you reach Cave #16. If you have not yet found him by that point, then you know your initial assumption that he started in an even-numbered cave is wrong.

That implies he must have started in an odd-numbered cave, such that on the following day he must be in an even-numbered cave. Thus, you can re-check Cave #16. If he is not there, on the next day he must be in an odd-numbered Cave #1-#15. The next day you check Cave #15, and so on down to Cave #2, at which point you must have caught him. Thus, the longest time it can take is #2, 3, 4, .... 15, 16, 16, 15, .... 3, 2, which adds up to 15 + 15 = 30 days.

On the first day, you check Cave #2. If he is in Cave #2, you'll catch him on the first day. If he's not in Cave #2 (but he did start in an even-numbered cave), then on the next day he must have moved into an odd-numbered cave between #3-#17 (since the only way he could get to #1 is to start out in #2, which you checked). So, on the next day, you can try Cave #3. If he's not there, then on the next day he must be in an even-numbered cave between #4-#16. Thus, you check Cave #4. In this manner, you keep checking adjacent caves until you reach Cave #16. If you have not yet found him by that point, then you know your initial assumption that he started in an even-numbered cave is wrong.

That implies he must have started in an odd-numbered cave, such that on the following day he must be in an even-numbered cave. Thus, you can re-check Cave #16. If he is not there, on the next day he must be in an odd-numbered Cave #1-#15. The next day you check Cave #15, and so on down to Cave #2, at which point you must have caught him. Thus, the longest time it can take is #2, 3, 4, .... 15, 16, 16, 15, .... 3, 2, which adds up to 15 + 15 = 30 days.

## 7. Fun with Combinatorics!

Simplify the following: (x-a)(x-b)(x-c)...(x-z).

**One of the terms is x-x = 0, so the answer is 0 :)**

loved the puzzles--I passed them along to some of my friends. Thank you for sharing

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