## Math puzzles for dinner [closed]

You're hanging out with a bunch of other mathematicians - you go out to dinner, you're on the train, you're at a department tea, et cetera. Someone says something like "A group of 100 people at a party are each receive hats with different prime numbers and ..." For the next few minutes everyone has fun solving the problem together.

I love puzzles like that. But there's a problem -- I running into the same puzzles over and over. But there must be lots of great problems I've never run into. So I'd like to hear problems that other people have enjoyed, and hopefully everyone will learn some new ones.

So: What are your favorite dinner conversation math puzzles?

I don't want to provide hard guidelines. But I'm generally interested in problems that are mathematical and not just logic puzzles. They shouldn't require written calculations or a convoluted answer. And they should be fun - with some sort of cute step, aha moment, or other satisfying twist. I'd prefer to keep things pretty elementary, but a cool problem requiring a little background is a-okay.

If you post the answer, please obfuscate it with something like rot13. Don't spoil the fun for everyone else.

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Since I see this has accumulated a couple of votes to close, I've started a meta thread: meta.mathoverflow.net/discussion/471/… – Anton Geraschenko Jun 24 2010 at 16:00
It's easy to forget the question, read one of the problems below, then write down an answer... – Gerald Edgar Jun 25 2010 at 0:28
Out of curiosity, is there a way of posting hidden text in answers that can be revealed by clicking on "Hidden Text?" (kind of like on Art of Problem Solving forums). – Alex R. Jun 26 2010 at 1:53
I find it a bit odd that there is a bounty on a CW. Can we discuss this [on meta](meta.mathoverflow.net/discussion/550/…)? – Willie Wong Jul 25 2010 at 14:46
Am I the only person who goes to social events with mathematicians and drinks, banters and has pointless debates about politics or films? – Yemon Choi Feb 2 2011 at 3:02

## closed as no longer relevant by Yemon Choi, Mark Meckes, Andres Caicedo, Scott Morrison♦Feb 2 2011 at 3:46

I really like the following puzzle, called the blue-eyed islanders problem, taken from Professor Tao's blog :

"There is an island upon which a tribe resides. The tribe consists of 1000 people, with various eye colours. Yet, their religion forbids them to know their own eye color, or even to discuss the topic; thus, each resident can (and does) see the eye colors of all other residents, but has no way of discovering his or her own (there are no reflective surfaces). If a tribesperson does discover his or her own eye color, then their religion compels them to commit ritual suicide at noon the following day in the village square for all to witness. All the tribespeople are highly logical and devout, and they all know that each other is also highly logical and devout (and they all know that they all know that each other is highly logical and devout, and so forth).

Of the 1000 islanders, it turns out that 100 of them have blue eyes and 900 of them have brown eyes, although the islanders are not initially aware of these statistics (each of them can of course only see 999 of the 1000 tribespeople).

One day, a blue-eyed foreigner visits to the island and wins the complete trust of the tribe.

One evening, he addresses the entire tribe to thank them for their hospitality.

However, not knowing the customs, the foreigner makes the mistake of mentioning eye color in his address, remarking “how unusual it is to see another blue-eyed person like myself in this region of the world”.

What effect, if anything, does this faux pas have on the tribe?"

For those of you interested, there is a huge discussion of the problem at http://terrytao.wordpress.com/2008/02/05/the-blue-eyed-islanders-puzzle/

Malik

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I have heard the following, politically less correct, version of this: A cruel custom on an island demands that every husband kills his wive at midnight if informed that she has cheated on him. All inhabitants of the island are married and all cheat. A newly arrived priest has heard of this scandalous lifestyle and, during his sermon, announces to the whole tribe that at least one (and thus at least two) inhabitant has been unfaithful. – Roland Bacher Jun 24 2010 at 14:13
The wikipedia link in the blog post is quite a revelation. – Nate Eldredge Jun 24 2010 at 19:28
I remember the blue-eyed islanders puzzle fondly from my teenage years, so about 20 years ago now. I am pretty sure it is not due to Prof. Tao. In fact I wonder is its provenance can be tracked down. (A version appears early on in Spivak's Calculus, for instance...) – Pete L. Clark Jun 26 2010 at 17:29
@Anixx: Well that is the interesting thing, and why this problem is popular. At first glance "no effect" seems to be right, but after some thought, we can deduce that all the blue eyed people kill themselves after a some number of days. – Eric Naslund Mar 6 2011 at 23:55
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There are some dwarves approaching to a bridge. They have to cross it to come back home from the cave where they work. Unfortunately, a dragon has just decided to reside under that bridge, and it's hungry. But it's also bored, so it doesn't want just to eat the dwarves, but proposes them a game: it will put on each of them one hat, either black or white, in no specific proportion (for example, it can happen all hats to be white). Of course they can't see their own hat, but they can see the others'. They will be then queueing at the beginning of the bridge, and each of them can just say one word. If this word matches with the colour of the hat that dwarf is wearing, then he's allowed to pass and to come back home. Otherwise, he'll be eaten by the dragon. Of course, they can decide for a strategy before the game begins.

What's the best strategy, and how many dwarves die on average?

EDIT: (deleted the previous PS, modified into this one) PS: Since I didn't solve all previous puzzles posted, I would be glad if someone could point me at equivalent puzzles, if any!

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A table with three legs does not wobble.

How about a quadratic table with four legs? Can it be rotated to fix the wobbling?

(Please assume reasonable conditions on life the universe and everything.)

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There was a puzzle in the Journal of Recreational Mathematics. I apologize if I am telling it incorrectly. A wire is stretched between two telephone poles. A flock of crows lands simultaneously on the wire. When they land, each crow looks at his nearest neighbor. What percentage of the flock are looking at a crow that is looking back at it?

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This problem appeared in the Australian Mathematical Society Gazette 35 (July 2008) page 151: "An odd number of wombats are standing in a field so that their pairwise distances are distinct. If each wombat is watching the closest other wombat to them, show that there is at least one wombat who is not being watched." Maybe this is the problem OP is trying to remember. – Gerry Myerson Jan 27 2011 at 4:37
It can't be independent of the number of crows: For n=2 it is 100% and for n=3 it is 2/3. – Sune Jakobsen Feb 1 2011 at 17:07

When you watch yourself in a mirror, left and right are exchanged. But why aren't top and bottom?

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I am used to even more misdirection: Describe it as a philosophy problem. In college, we would ask philosophy majors this puzzle at dinner, and let them go on at length before revealing that there is a simple answer, which shouldn't be the case for a philosophy problem. – Douglas Zare Jan 28 2011 at 10:02
Gur zveebe qbrf abg rkpunatr yrsg naq evtug, abe qbrf vg rkpunatr gbc naq obggbz; vg rkpunatrf sebag naq onpx. Zbfg crbcyr, jura gheavat nebhaq gb ybbx va n zveebe, jvyy ghea nobhg n iregvpny nkvf engure guna n ubevmbagny bar. Guvf gheavat unf gur rssrpg bs rkpunatvat sebag naq onpx (chggvat gurz onpx gur jnl gurl jrer) nf jryy nf yrsg naq evtug. – Tanner L. Swett Mar 31 2012 at 14:36
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Saw this recently:

There is a board with a grid drawn on it. You hammer a few nails into the board at intersection points and then stretch a rubber band around the nails. Then you observe that

(1) you can't take away any of the nails without changing the shape,

(2) the rubber band does not enclose (or pass over) any grid point without a nail in it, and

(3) you can't add another nail and extend the rubber band around it without (1) or (2) becoming untrue.

How many nails are there?

(The answer is obvious and easy to show in a few lines, but I like it because it's quicker to figure out the corresponding problem in d dimensions and then see what pops out for d=2 than to make a specific two-dimensional argument that isn't the generic argument with d=2.)

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Can one partition the plane $\mathbb{R}^2$ by closed intervals of equal length? How? The answer to the first question is "yes". In other words, can one cover the plane with translates and rotations of a given closed line segment such that every point lies on exactly one segment?

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Let $I$ be the set of irrational numbers, with the usual topology. Is $I$ homeomorphic to $I\times I$?

Edited to add: In fact, it's an even better puzzle if you replace $I$ with $Q$, the rational numbers.

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Q is the unique (up to homeomorphism) countable metric space without isolated points. Q x Q is also such a space so Q x Q == Q. P (the irrationals) is the unique (up to homeomorphism) completely metrizable zero-dimensional separable space that has no compact sets with non-empty interior ("nowhere locally compact"). P x P is one too... (and so is N^N, which is homeomorphic to P via this theorem or via the map using continued fractions. And N^N x N^N == N^(N+N) == N^N) – Henno Brandsma Dec 8 2010 at 8:33

Quite simply, a monkey's mother is twice as old as the monkey will be when the monkey's father is twice as old as the monkey will be when the monkey's mother is less by the difference in ages between the monkey's mother and the monkey's father than three times as old as the monkey will be when the monkey's father is one year less than twelve times as old as the monkey is when the monkey's mother is eight times the age of the monkey, notwithstanding the fact that when the monkey is as old as the monkey's mother will be when the difference in ages between the monkey and the monkey's father is less than the age of the monkey's mother by twice the difference in ages between the monkey's mother and the monkey's father, the monkey's mother will be five times as old as the monkey will be when the monkey's father is one year more than ten times as old as the monkey is when the monkey is less by four years than one seventh of the combined ages of the monkey's mother
and the monkey's father.

If, in a number of years equal to the number of times a monkey's mother is as old as the monkey, the monkey's father will be as many times as old as the monkey as the monkey is now, find their respective ages.

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@Yaakov: That's right; however, it wasn't until I started memorizing the answer that I stopped getting invites. – ThudnBlunder Nov 30 2010 at 12:28

I understand your feeling , I myself know lots of them . Among original ones

http://www.amazon.com/Mathematical-Puzzles-Connoisseurs-Peter-Winkler/dp/1568812019

This Peter Winkler does something that is rarely done and is a must not only for a mathematician but for a connoisseur: He produces declination of a problem.

In fact there are two books of his.

Another source of problems that you may like is "IBM ponder this" AT

http://domino.research.ibm.com/Comm/wwwr_ponder.nsf/pages/index.html

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Yes, I suggested the Winkler books in my answer of 24 June. – Gerry Myerson Oct 1 2010 at 12:34

You and your adversary have a sufficiently large bag of identical coins, and are seated on opposite sides of a rectangular table. You take turns placing coins on the table. The first one that cannot put a coin on the table without overlapping any other coin loses. What is your strategy to always win if you're allowed to start?

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Here's one that I like that I just heard a few days ago. Alice and Bob play the following game. Alice is randomly dealt 5 cards from an ordinary deck of cards. She is allowed to show Bob 4 of the 5 cards (in order). Bob must then guess what the 5th card is.

Prove that Alice and Bob have a strategy where Bob can always guess correctly.

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I'll just remark that the strategy is constructive in the sense that two non-mathies could easily execute it. – Tony Huynh Aug 10 2010 at 17:49
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I think this has not been published yet. Apologies otherwise. I learnt it from Antonio Sánchez Calle in my first year of undergraduate and I had 3 non-mathematicians thinking about it for about 4 hours, so there is a guaranteed success if you tell around :)

5 people are shipwrecked in a deserted island. They find a monkey and lots of coconuts. They spend the whole day collecting coconuts that they keep together and since they are tired they go to sleep. The first person wakes up, attempts to divide the amount of coconuts in five parts, but one of them is spared, so he gives to the monkey. Then he eats one fifth of the coconuts and goes back to sleep.

The second person wakes up and follows the same procedure. He divides the coconuts in five, one is spared and he gives it to the monkey, eats his share and goes back to sleep.

The third, forth and fifth people do the same thing. How many coconuts were there at the beginning? (modulo something, of course).

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-4 (modulo 5^5), i.e. 3121 for non-mathematicians... – Andrei Moroianu Nov 16 2010 at 22:13
I actually haven't computed it since I was in my first year, but I remember that was the answer :) Problems with monkeys are happier problems. Monkeys with problems are sadder monkeys. – Jesus Martinez Garcia Nov 18 2010 at 9:19

Simple puzzles; unfortunately, I do not know how to formulate them in a whimsical fashion suitable for a dinner, they very much sound like math puzzles.

1. Take n labeled points $x_1, \dots, x_n$ in the plane. How do you construct a n-gon $a1, \dots, an$ such that for all i, $x_i$ is the midpoint of $[a_i, a_{i+1}]$ (with the convention $a_{n+1}=a_1$ of course). I was surprised to come across this problem in the puzzle pages of Le Monde. I think non-mathematicians would have a hard time with it.

2. For mathematicians who don't already know it, the Sylvester Gallai Theorem can offer stimulating after dinner discussions (or during those long proctoring sessions).

3. A napkin should be enough for this one (if even needed!). Consider a map f from the plane to the reals such that the sum of the values of f on the vertices of any square is zero. Find all such maps.

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Here's one of my favorites. There are 99 bags, each of which contains some number of apples and some number of oranges. Prove that there's a way to select 50 out of the 99 bags, such that these 50 simultaneously contain at least half the total number of apples and at least half the total number of oranges.

One fun aspect of this problem is that there are a number of distinct solutions, inspired by different areas of math. I know of at least three...

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Good point, BlueRaja. Problem edited accordingly. (The simplest counterexample to the original formulation is if all the bags are empty.) But, if you require that all bags have a <i>strictly positive</i> number of each fruit, then you can also get a strict inequality. – Dave Futer Aug 6 2010 at 21:17

There is a plane with 100 seats and we have 100 passengers entering the plane one after the other. The first one cannot find his ticket, so chooses a random (uniformly) seat. All the other passengers do the following when entering the plane (they have their tickets). If the seat written on the ticket is free, one sits on this seat, if not he chooses a other (free) seat at random (uniformly). What is the probability the last passenger entering the plane gets the correct seat?

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You have a large pile of ropes and some matches. All you know about the ropes:

• Each rope has a different length
• Each rope burns completely (starting from one end) in exactly 64 minutes
• Each rope has non-uniform density, meaning it is thicker at some points than others. Consequently, burning half a rope cannot be guaranteed to take 32 minutes.

The goal is to identify when exactly 63 minutes have passed.

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Hint: lbh arrq gb ohea frira ebcrf – BlueRaja Aug 6 2010 at 17:58

A puzzle(rather, a tale to lure the reader into the domain of complex numbers) lifted from George Gamow's "One, Two, Three, Infinity":

There was a young and adventurous man who found among his great-grandfather’s papers a piece of torn parchment that revealed the precise location of a hidden treasure. The instruction reads:

Sail to North latitude __ and West longitude __ where thou wilt find a deserted island. There lieth a large meadow, not pent, on the north shore of the island where standeth a lonely oak and a lonely pine tree. There thou wilt see also an old gallows on which we once were wont to hang traitors. Start thou from the gallows and walk to the oak counting thy steps. At the oak thou must turn right by a right angle and take the same number of steps. Put here a spike in the ground. Now must thou return to the gallows and walk to the pine counting thy steps. At the pine thou must turn left by a right angle and see that thou takest the same number of steps, and put another spike into the ground. Dig halfway between the spikes; the treasure is there.

The instructions being quite clear and explicit, our young man chartered a ship and sailed to the South Seas. He found the island, the field, the oak and the pine, but to his great sorrow, the gallows was gone. Too long a time had passed: rain and sun and wind had disintegrated the wood and returned it to the soil, leaving no trace of the place where once it had stood. Our adventurous man fell into despair. Digging all over the field at random, he found nothing and sailed back empty-handed.

A sad story for sure, but sadder to think that he might have easily located the treasure had he known a little about the arithmetic of complex numbers!!

Question: How???

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Nice. Instead of making random guesses about the location of the treasure, he should have made at least one random guess about the former location of the gallows! – Tracy Hall Aug 6 2010 at 21:43
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Here's one I saw a while ago:

A prisoner is presented with the following challenge by one of the guards of the jail. The prisoner is to be blindfolded and then the guard will place $n$ coins on a circular turntable with any combination of heads and tails facing up (with at least one tails showing initially). The prisoners goal is to flip over coins until all heads are showing.

This would be easy enough if the guard did not interfere. The prisoner could just try all $2^n$ combinations, and one of them would be guaranteed to result in all heads. However, to complicate matters, the guard may turn the table during this process. More specifically, the following process is repeated. First, the prisoner chooses a set of positions of coins to flip over. Then, before the coins are flipped, the guard turns the turntable so as to try to prevent the prisoner from flipping all of the coins to heads. Finally, the prisoner flips over the coins that are at the positions chosen in the first step. If all heads are showing, the game stops and the prisoner is set free.

The question is, for what values of $n$ does the prisoner have a winning strategy and how many moves does it take?

What if the guard uses 6-sided dice instead of coins with the goal of showing all ones (assuming the orientations of the dice are preserved relative to their positions on the turntable between rounds)?

In general, what values of $n$ allow a solution with $k$-sided dice?

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BlueRaja, the guard can turn the table maliciously. The prisoner chooses the locations to flip, and then the guard can turn the table any amount in an effort to prevent the coins from showing all heads. – jonderry Aug 26 2010 at 22:23

A magician places $N = 64$ coins in a row on a table, then leaves the room. A person from the audience is then asked to flip each coin however he likes [so there are $2^{64}$ possible states]. He is also asked to mention a number between 1 and $N$. After this, the magician's assistant flips exactly one coin. The magician reenters the room, looks at the coins and "guesses" the number chosen by the audience.

What is their strategy? For which values of $N$ can the trick be performed?

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You are the captain of a team of N players, in charge of choosing a strategy that your adversary will overhear (and therefore rig the game for you to lose unless the strategy is perfect). To play the game, the adversary writes a distinct name on each player's forehead and you are brought into a situation where each of you can learn the name given to every other player, but not your own. Naturally you cannot communicate once the game has started. Each of you is blindfolded and given a single invertible glove. On a signal, each of you silently places your glove on one hand or the other. You are then lined up in alphabetical order by the names on your foreheads, all facing the same direction, and you join hands in one long chain. If any of you touches another player's glove with your bare hand the team loses, but if it is always hand-to-hand and glove-to-glove, you are victorious.

For what values of N can you give your team a winning strategy, and what is it?

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Here is a classic:

Plant 10 trees in five rows, with 4 trees in each row.

I like this because there are two basic approaches to the problem: the one almost everyone thinks of and uses to grind slowly towards a solution, and the one they should think of instead, which leads quickly to many solutions.

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Okay, I've got one, and as far as I know it hasn't been analyzed before.

I was watching a travel show the other night -- they were in Korea, and a group of people were playing a drinking game. It works like this:

One person is "it". This person says something like, "ready, set..." then points at one other player and calls out a number between 2 and n (where n is the number of people playing). At the same time, everyone else also points at one other player. Then, for whatever number got call out, you jump that many steps from the "it" person, and that person has to drink. So if I call out "two" and point at Joe, and Joe points at Bob, then Bob has to drink.

I think the game is pretty interesting, mathematically, especially when you allow numbers greater than n to be called. One interesting thing I found: with n=3, if you call 7 (or 7+6x, where x is a non-negative integer), you are guaranteed to stick the player you initially point at, no matter who points to whom.

I think an interesting question is, given n players, what is the smallest number the 'it' person can call that guarantees he will not stick himself? (I have an answer, but I want to see if you all come up with the same thing. :-) And what's the best strategy for the caller if you enforce the rule that you must call out a number between 2 and n? What's the best strategy for the other players, if they're allowed to collude on who they're going to point to? Etc.

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Here's a balance scale problem that I decided to post because a little bit of googling around for it came up negative. It differs from most balance scale puzzles I've seen because it doesn't involve "bad weights". I learned of it from a friend of mine who is an engineer.

There are 10 balls which come in two possible weights. Using a balance scale at most 3 times, determine whether all the balls are the same weight or not.

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It is very important that you tell these two puzzles in the correct order, i.e., first the first puzzle and then the second one. The first puzzle is very easy but messes with people's minds in just the right way. In my experience some mathematicians are driven crazy by the second puzzle.

Puzzle 1: Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She wanted to divide the cake fairly among her 9 grandchildren. How should she cut the cake?

Puzzle 2: Grandma made a cake whose base was a square of size 30 by 30 cm and the height was 10 cm. She put chocolate icing on top of the cake and on the sides, but not on the bottom. She wanted to divide the cake fairly among her 9 grandchildren so that each child would get an equal amount of the cake and the icing. How should she cut the cake?

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uneq sbe gur zngurzngvpvnaf, ohg ernyyl rnfl sbe gur tenaqzn... – Yaakov Baruch Jul 7 2010 at 21:57
@ BlueRaja: gurer ner ab gevpxf, vg ernyyl vf n fvzcyr naq ryrzragnel 2Q trbzrgel ceboyrz. Nyfb abgr gung vg znggref gung gur pnxr vf fdhner naq abg erpgnathyne. – Yaakov Baruch Jul 9 2010 at 10:29
Hint: gur fvzcyr fbyhgvba (jryy, ng yrnfg gur bar V'z guvaxvat bs) trarenyvfrf vzzrqvngryl gb nal ahzore bs tenaqpuvyqera (jvgubhg punatvat gur pnxr). – Peter LeFanu Lumsdaine Jul 12 2010 at 17:58
@Ryan: no, no, can't remove the icing of course. – Andrej Bauer Jul 27 2010 at 6:22
Solution, since nobody posted it: Gb qvivqr gur pnxr vagb a cnegf, znex a cbvagf ba gur pvephzsrerapr bs gur fdhner fhpu gung gurl qvivqr gur pvephzsrerapr vagb rdhnyyl ybat cnegf. Gura phg sebz gur pragre bs gur fdhner gb gur znexrq cbvagf. – Andrej Bauer Jul 27 2010 at 23:27

Okay, so it's somewhat more numeric than the others, but I quite enjoy the simplicity of:

### Simplify:

$$\sqrt{2+\sqrt3}$$

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(GURER FUBHYQ OR AB VGRENGRQ FDHNER EBBGF VA LBHE NAFJRE) – Tom Boardman Jul 6 2010 at 12:30
Nalbar jub qvq zngu pbzcrgvgvbaf erpragyl fubhyq erpbtavmr gur unys-natyr sbezhyn, V jbhyq guvax. :) – JBL Jul 6 2010 at 12:44
Jung'f guvf gevt vqragvgl ohfvarff? Chg n zvahf orsber gur vaare fdhner ebbg naq lbh trg gur vairefr. Gura fhz be fhogenpg vg jvgu vgf vairefr naq frr jung gurve fdhnerf ner. Gur tnybvf gurbel fbyhgvba :) – Ryan Reich Jul 28 2010 at 4:21

First, I apologize that this puzzle is not "clean". It's more suitable for the bar rather than dinner (in fact I heard it at a party with mathematicians). I understand if it should be taken down or rephrased. I scanned the previous puzzles and I don't think this puzzle is a duplicate.

Suppose you are male(female) and stranded on an island with three females(males). You wish to have protected sex with all three females(males), but you only have two condoms and no other forms of protection. Clearly you can have protected sex with two of them by using one condom, throwing it away and using the other. Can you have protected sex with all three of them?

By protected I mean there is no exchange of fluids from one person to another, i.e. you can't use a condom with one person and then use it "as is" with another person. Also, despite being surrounded by the ocean you cannot just rinse them off - that's dirty.

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You can save yourself the bother of having to write "male (female)" all the time AND make the statement of your puzzle more inclusive by avoiding any mention of gender. Just saying. – David Steinberg Jul 6 2010 at 5:45
The puzzle can be stated in a form suitable for a general audience, e.g., a doctor wants to perform surgery on three patients but has only two sets of surgical gloves, etc. – Gerry Myerson Jul 6 2010 at 5:47
We've had this puzzle before on MO, haven't we? – Yemon Choi Jul 6 2010 at 6:01

This is a hat problem I heard only two days ago: you have a hundred people and each one has a (natural) number between 1 and 100 written on his hat. (Numbers may repeat.) As ususal, everybody can see only the numbers on other people's hats. Give these guys a strategy for guessing so that at least one will surely make the right guess. (They do not hear each other guesses.)

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Here is another of my favorites: Player 1 thinks of a polynomial P with coefficients that are natural numbers. Player 2 has to guess this polynomial by asking only evaluations at natural numbers (so one can not ask for $P(\pi)$). How many questions does the second player need to ask to determine P?

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Qbrf Cynlre 2 xabj n obhaq P sbe gur pbrssvpvragf? (Gura n fvatyr k1>P jvyy erirny nyy bs gurz ol jevgvat C(k1) va onfr k1.) – Yaakov Baruch Jun 29 2010 at 8:27
nu... abj vg qnjarq ba zr. C(1) jvyy rfgnoyvfu n obhaq ba gur pbrssvpvragf, gb or hfrq jvgu cerivbhf nggrzcgrq nafjre. – Yaakov Baruch Jun 29 2010 at 11:15
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What is the resistance between 2 adjacent vertices of an infinite checkerboard if every edge is a 1 ohm resistor?

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xkcd.com/356 – Chris Phan Jul 1 2010 at 15:14
Great puzzle, but having seen the solution I'm not sure I'd have managed this over dinner... using a ynggvpr terra'f shapgvba hardly puts this at the level of sudoku does it? – Tom Boardman Jul 13 2010 at 11:29