# 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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## closed as no longer relevant by Yemon Choi, Mark Meckes, Andrés Caicedo, Scott Morrison♦Feb 2 '11 at 3:46

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Since I see this has accumulated a couple of votes to close, I've started a meta thread: tea.mathoverflow.net/discussion/471/math-puzzles-for-dinner – Anton Geraschenko Jun 24 '10 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 '10 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 '10 at 1:53
I find it a bit odd that there is a bounty on a CW. Can we discuss this on meta? – Willie Wong Jul 25 '10 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 '11 at 3:02

Given three equal sticks, and some thread, is it possible to make a rigid object in such a way that the three sticks do not touch each other? (all objects are 1 dimensional; sticks are straight and rigid, and the thread is inestensible).

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Suppose 100 ants are placed randomly (with random orientations) along a yard stick. Each ant walks at a pace of an inch a minute. Each time two ants meet, they instantaneously reverse direction, and if an ant meets the end of the yardstick, it instantaneously reverses direction.
Do the ants ever return to their starting positions? At what time? (A yardstick is 36 inches long.)

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A closely related puzzle: gilkalai.wordpress.com/2008/10/23/two-math-riddles – Gil Kalai Jun 26 '10 at 18:29
Also, must the ants also return to their starting orientations? – Qiaochu Yuan Jun 26 '10 at 19:48

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 '10 at 21:17

How many times a day is it impossible to tell the time by a clock with identical hour and minute hands, provided you can always distinguish between a.m and p.m? P.S. Ask them for a fast answer.

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Taken from the book "The art of mathematics; Coffee time in Memphis" by B. Bollobas. Some of the puzzles there need a paper though.. – Gjergji Zaimi Jun 24 '10 at 5:55

I had a good outcome with this one once. It probably helped that the other two mathematicians had a drink or two before dinner. (Otherwise they would have solved it in 5 seconds...) When salad was served, somebody had oil and vinegar in separate little pitchers...

Suppose you have two containers, one with oil, one with vinegar, equal volume. Take one teaspoon of the oil, put it into the vinegar, stir. Than take one teaspoon of the mixture, put it into the oil, stir. Now: is there more vinegar in the oil or more oil in the vinegar?

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This is the same as one of Douglas S. Stones's puzzles. – j.c. Jun 24 '10 at 18:50
...that I heard from someone at dinner, and presumably he heard from someone at dinner, etc. – Douglas S. Stones Jun 25 '10 at 5:05

Alice secretly picks two different real numbers by an unknown process and puts them in two (abstract) envelopes. Bob chooses one of the two envelopes randomly (with a fair coin toss), and shows you the number in that envelope. You must now guess whether the number in the other, closed envelope is larger or smaller than the one you’ve seen.

Is there a strategy which gives you a better than 50% chance of guessing correctly, no matter what procedure Alice used to pick her numbers?

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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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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 '10 at 21:43

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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Start with four beads placed at the corners of a square. You are allowed to move a bead from position x to position y if one of the other three beads is at position (x+y)/2. In other words, you may `reflect a bead with respect to another bead.' Find a sequence of such moves that places the beads at the corners of a bigger square, or show that the task is impossible.

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Here's an easy but fun one (probably suitable for a class or for mathematicians who have had some drinks). You have ten bags of coins, one of which contains fake coins. We may of course assume that each bag contains infinitely many coins. The real coins weigh 1 gram each, while the fake coins weigh .9 grams each. You have a scale, which is capable of only one accurate reading before breaking. Determine which bag contains the fake coins.

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Avpr bar - gnxvat 10 pbvaf sebz gur svefg, avar sebz gur frpbaq... bar sebz gur ynfg tvirf gur nafjre, evtug? – Harun Šiljak Jun 25 '10 at 9:57

Cryptography riddle - that's a branch of mathematics, right? :)

Jan and Maria have fallen in love (via the internet) and Jan wishes to mail her a ring. Unfortunately, they live in the country of Kleptopia, where anything sent through the mail will be stolen unless it is enclosed in a padlocked box. Jan and Maria each have plenty of padlocks, but none to which the other has a key. How can Jan get the ring safely into Maria’s hands?

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Good thing SOMEONE solved this problem, otherwise we wouldn't have email! – Dylan Wilson Jun 25 '10 at 4:36
@Dylan: Actually, email is not normally encrypted or authenticated - you have to go through great pains (en.wikipedia.org/wiki/Pretty_Good_Privacy) for that. It's possible for anyone listening to your network traffic to read your emails, and for anyone to send email as you! – BlueRaja Jun 25 '10 at 15:36
Do not ever assume that the sender of the email is the address written in the From: field. It is the same as assuming that the sender on the envelope of a letter you received is correct. – Andrea Ferretti Jun 25 '10 at 16:05
Guvf bar vf snveyl snzbhf: Wna fraqf gur evat va n ybpxrq obk. Znevn nqqf n frpbaq ybpx (ure bja) gb gur fnzr obk, naq fraqf vg onpx. Wna erzbirf uvf cnqybpx, naq fraqf vg gb Znevn, jub pna bcra vg gb ergevrir gur evat. – JBL Jun 28 '10 at 1:33

An evil sorceress is holding 100 princes captive. Right now they are all in the same prison cell and can discuss strategy. However, in a moment, they will each be taken to individual prison cells, where no communication is possible. After that, the sorceress will start randomly calling princes to her bedroom (one at a time). This continues indefinitely, so a prince can visit the bedroom many times. The bedroom has two light switches, whose state can be observed only from inside the bedroom. When a prince is called to the bedroom, he can observe the state of the switches, and then must change the state of exactly one of the light switches. The initial state of the light switches is not known.

The princes will be set free if any one of them can determine if all of them have been called to the room.

Puzzle: Determine a strategy for the princes so that they are guaranteed to be set free eventually. The strategy should never output a false positive. For example, if a prince has been called one million times he can reason that on average, everyone else has been called one million times. Thus it is very likely that all the princes have been called to the bedroom, but it is possible that one prince still hasn't been called.

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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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Take a convex polyhedron and any point inside of it. To every face, drop a normal line from the point. Note that it is both possible to land inside the face or outside. Construct a polyhedron where every such normal line drops outside of the corresponding face, or prove such a polyhedron cannot exist.

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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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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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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 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 '10 at 17:49

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 '10 at 22:23

What is four thousand and ninety-nine plus one?

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I don't know why this one was voted down; maybe someone was too clever. The point is that if you ask someone this quickly their first response is likely to be five thousand. – Qiaochu Yuan Jun 26 '10 at 8:32
I actually did say "five thousand" before reading a bit closer. – Michael Lugo Jun 26 '10 at 12:50
Qiaochu - If that's the case then it's not so much a puzzle as an attempted trick. – Mark Jun 27 '10 at 6:41

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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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 '10 at 8:33

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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Assuming you have unlimited time and cash, is there a strategy that's guaranteed to win at roulette?

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Qbhoyvat gur fgnxr nsgre rirel ybff? Gung jnl lbh nyjnlf jva n qbyyne :) – Harun Šiljak Jun 25 '10 at 9:58
Shaal, zl nafjre jnf "ohl gur pnfvab"...gubhtu V thrff vg'f abg va gur fcvevg – Charles Siegel Jun 25 '10 at 19:58

You're on a path on an island, come to a fork in the road. Both paths lead to villages of natives; the entire village either always tells the truth or always lies (both villages could be truth-telling or lying villages, or one of each). There are two natives at the fork - they could both be from the same village, or from different villages (so both could be truth-tellers, both liars, or one of each).

One path leads to safety, the other to doom. You're allowed to ask only one question to each native to figure out which path is which.

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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 '10 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 '10 at 5:47
We've had this puzzle before on MO, haven't we? – Yemon Choi Jul 6 '10 at 6:01

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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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 '10 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 '10 at 9:19

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.