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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.

One problem per answer.

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

Charles Matthews
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Richard Dore
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67 Answers67

54

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

Malik Younsi
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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 '10 at 14:13
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    The wikipedia link in the blog post is quite a revelation. – Nate Eldredge Jun 24 '10 at 19:28
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    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 '10 at 17:29
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    What's the point here? Obviously, no effect: the islanders already know that there are many people with blue eyes on the island. Even if they do not know statistics, they know that such people exist there. – Anixx Jan 28 '11 at 14:11
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    @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 06 '11 at 23:55
  • Just passed by this and remembered waitbutwhy blog posting another version of the same puzzle. – polfosol Mar 07 '21 at 21:23
46

You and infinitely many other people are wearing hats. Each hat is either red or blue. Every person can see every other person's hat color, but cannot see his/her own hat color; aside from that, you cannot share any information (but you are allowed to agree on a strategy before any of the hats appear on your heads). Everybody simultaneously guesses the color of his/her hat. You win if all but finitely many of you are right. Find a strategy so that you always win.

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    As I recall, the number of hat colors can be arbitrary...? I guess it's better to state this version because if it sounds too impossible people will get suspicious and might catch on faster. – Qiaochu Yuan Jun 24 '10 at 05:12
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    @Qiaochu: yes, that's true. A variant where it's important to have two colors: you win if everybody is right or everybody is wrong. – Anton Geraschenko Jun 24 '10 at 05:15
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    There's a solution to this with countably many people and uncountably many possible hat colors. – Charles Chen Jun 24 '10 at 05:27
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    ...and in fact it's sufficient that everybody can see all but finitely many hats. – Pietro Majer Jun 24 '10 at 06:18
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    The solution to this problem can be turned into a strategy to predict the future (!?). It was nicely explained in a nice article on the American Mathematical Monthly, of which I forgot the title. Sadly the strategy is not effective... – Andrea Ferretti Jun 24 '10 at 13:29
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    @Andrea: The name of the article is A Peculiar Connection Between the Axiom of Choice and Predicting the Future by Hardin and Taylor. You can access the article by clicking on the link from Francois G. Dorais' answer to this question – Tony Huynh Jun 24 '10 at 15:07
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    As closing the question is being debated, I would point to this particular answer as a factor against it. – Yaakov Baruch Jun 24 '10 at 18:15
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    This is one of my favorites. I've always liked the mental image of the players sitting around beforehand and nterrvat ba n pubvpr shapgvba. – Nate Eldredge Jun 24 '10 at 19:13
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    This is probably my favorite puzzle. Anybody who hasn't seen this before should really work it out; it's a great one. – Michael Burge Jun 25 '10 at 14:42
  • @AntonGeraschenko Even in the everybody right/wrong variant, there is a strategy for any cardinality of colors. – Elliot Glazer Oct 23 '19 at 02:07
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You are blindfolded, then given a deck of cards in which 23 of the cards have been flipped up, then inserted into the deck randomly (you know this). Without taking the blindfold off, rearrange the deck into two stacks such that both stacks have the same number of up-flipped cards. (You are allowed to flip as many cards as you please.)

Vectornaut
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Kiochi
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    I really like this one. I heard it a few years ago, and remembered my solution because it was neat. However, I had forgotten the question (and have been intermittently searching the internet for it) until now. Thanks!

    (It is a bad thing to have an answer without a question floating around in your head!)

    – ADL Jun 24 '10 at 14:40
  • You mean each stack contains exactly $11.5$ cards which are up-flipped? Am I allowed to tore every card into two equal pieces? – Roland Bacher Jun 24 '10 at 16:17
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    @Roland: Each stack contains the same number of up-flipped cards, not necessarily 11.5 (in fact, definitely not, as ripping cards is destructive). – Kiochi 0 secs ago – Kiochi Jun 24 '10 at 16:53
  • Although I should mention that you are allowed to flip as many cards as you please. – Kiochi Jun 24 '10 at 16:54
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    @Kiochi: you really should mention this in the answer. I've edited it for you, hope that's okay! – Vectornaut Jun 24 '10 at 18:15
  • @Vectornaut: you mean in the question; yes I agree. Thanks for editing! – Kiochi Jun 24 '10 at 18:39
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    This thread is killing me. Won't anyone post the answers? – Olivier Jun 25 '10 at 08:20
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    Bar uvag: va gur fbyhgvba V xabj, lbh jvyy abg xabj ubj znal pneqf ner hc-syvccrq va rnpu cvyr, bayl gung gur ahzore bs fhpu vf rdhny. – Kiochi Jun 25 '10 at 14:06
  • Are you sure that 23 isn't supposed to be 26 (half of a deck)? – JBL Jun 27 '10 at 03:09
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    @JBL: Yes, we're sure. The answer is humiliatingly easy. – TonyK Jun 27 '10 at 22:35
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    Yes, you're right -- I just hadn't figured out how to adapt the solution for 26 to other numbers: Qvivqr gur qrpx vagb bar cvyr bs fvmr gjragl-guerr naq bar cvyr bs fvmr gjragl-avar. Abj ghea gur svefg cvyr hcfvqr-qbja. (Naq bs pbhefr 23 vf neovgenel urer.) – JBL Jun 28 '10 at 01:27
  • @Tonyk: I was confused until I read your hint. That was humiliatingly easy! – Eric Naslund Mar 07 '11 at 00:04
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1000 prisoners are in jail. There's a room with 1000 lockers, one for each prisoner. A jailer writes the name of each prisoner on a piece of paper and puts one in each locker (randomly, and not necessary in the locker corresponding to the name written on the paper!).

The game is the following. The prisoners are called one by one in the room with the lockers. Each of them can open 500 lockers. If a prisoner finds the locker which contains is name, the game continues meaning that he leaves the room (and leaves it is the exact same state as when it entered it, meaning that he cannot leave any hint), and the following prisoner is called. If anyone of the prisoners fails to recover his name, they all lose and get killed.

Of course they can agree before the beginning of the game on a common strategy, but after that, they cannot communicate anymore, and they cannot leave any hint to the following prisoners.

A trivial strategy where each prisoner opens 500 random lockers would lead to a winning probability of 1/2^1000. But there exists a strategy that offers a winning probability of roughly 30%.

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    This is also a very beautiful puzzle. As to the context, I'prefer to tell it in a less violent fashion: let's say that the director of a private prison offers a free Thanksgiving dinner to his clients if they succeed to win the game. On the mathematical side, there is a less known but very nice additional part: prove that the solution is indeed optimal: no other strategy gives better chances of winning. – Pietro Majer Jun 24 '10 at 17:59
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    This is a very nice puzzle. Of course, computing that winning probability requires some calculation, but finding the strategy is beautiful math. – Tara Holm Jun 25 '10 at 15:09
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    I love this puzzle too. I first read it on http://www.madore.org/~david/weblog/2006-11.html#d.2006-11-20.1390 which also lists a few more well-known puzzles. – Zsbán Ambrus Jun 25 '10 at 22:09
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    @Pietro: but that's half the fun, free dinner doesn't motivate a good strategy like the guillotine does. – Adam Hughes Dec 07 '10 at 23:59
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    This may be pedantic, and is definitely very late, but it is important that the lockers are labeled. If the prisoners are unable to tell different lockers apart then the (very pretty) solution doesn't work. – Peter Samuelson Jul 27 '11 at 02:51
40

There are $n$ balls rolling along a line in one direction and $k$ balls rolling along the same line in the opposite direction. The speeds of the balls in the first group and in the second group are equal. Initially the two groups of balls are separated from one another and at some point the balls start colliding. The collisions are assumed to be elastic. How many collisions will there be?

algori
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    This is a fun puzzle!

    V guvax vg vf gur cebqhpg bs gur ahzore bs onyyf zbivat gb gur yrsg gvzrf gur ahzore bs onyyf zbivat gb gur evtug. (V zbqry guvf hfvat fcva punvaf...)

    – José Figueroa-O'Farrill Jul 25 '10 at 02:34
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    Jose -- glad you liked it. Your answer is correct. Actually, fvapr gur pbyyvfvbaf ner rynfgvp, bar pna guvax bs gur onyyf nf vs gurl cnffrq evtug guebhtu rnpu bgure vafgrnq bs pbyyvqvat. – algori Jul 25 '10 at 03:12
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    I've always liked this puzzle, because most people eventually solve it, but without noticing the elegant solution. – Eric Tressler Jul 26 '10 at 22:50
  • I think you have to add something about the initial conditions. I can easily envisage an instance where there is exactly one collision. – Christian Blatter Jan 27 '11 at 13:08
  • Christian -- or none at all! You are right, thanks, will correct this. – algori Jan 27 '11 at 20:59
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(I learned this puzzle from Ravi Vakil.) Suppose you have an infinite grid of squares, and in each square there is an arrow, pointing in one of the 8 cardinal directions, with the condition that any two orthogonally adjacent arrows can differ by at most 45 degrees.

Can there be a closed cycle? (i.e. start at some arrow, move to the square that arrow points to, follow where the arrow there points and so on, and come back to the square you started at).

Henry Segerman
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29

Adam Hesterberg told me this one ages ago. It apparently used to circulate around MOP.

Three spiders and a fly are placed on the edges of a regular tetrahedron, and travel only on those edges. The fly travels at the rate of $1$ edge/s, whereas the spiders travel at the rate of $1 + \epsilon$ edge/s for some $\epsilon > 0$. The spiders want to agree beforehand on a deterministic strategy for capturing the fly, whose location they do not know (but they do know each others' locations). The fly is invisible and omniscient; in particular, it is aware of the locations of the spiders and of their strategy at all times. (It also cannot fly.)

Can the spiders guarantee that they will catch the fly in finite time, regardless of the initial positions of the spiders and the fly? Does the answer depend on the value of $\epsilon$?

Qiaochu Yuan
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    I like how the fly is anything but a fly.. :) – Gjergji Zaimi Jun 24 '10 at 05:46
  • I initially read that with the bugs being placed on the vertices, making it a bit stupid... – Harry Altman Jun 24 '10 at 06:53
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    Va pnfr nalbar vf phevbhf, gur nafjre vf lrf, ohg V unir sbetbggra gur fgengrtl. Gur fcvqref pna pngpu gur syl va fbzrguvat yvxr gjb bire rcfvyba cyhf guerr frpbaqf. – Qiaochu Yuan Jun 26 '10 at 02:36
  • I'd enjoy seeing a quick answer to this if anyone knows it. V'ir sbhaq n enaqbz fgengrtl gung jbexf cerggl jryy, ohg V nffhzr gur bar Dvnbpuh sbetbg vf qrgrezvavfgvp. – Jonah Ostroff Jun 26 '10 at 15:02
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    @Jonah: Gur syl vf bzavfpvrag, fb cerfhznoyl vg xabjf enaqbz pubvprf nurnq bs gvzr. Engure, fnl gur iregvprf ner pbybherq. Sebz fgneg cbfvgvbaf, gjb fcvqref zbir gb erq, bar gb juvgr. Gura: bar ubyqf ng erq, bar zbirf erq gb terra, bar zbirf juvgr gb terra; terra ubyq, terra gb oyhr, erq gb oyhr; oyhr ubyq, oyhr gb juvgr, terra gb juvgr; juvgr ubyq, juvgr gb erq, oyhr gb erq. Ercrng. Gur syl vf sbeprq vagb gur erq-terra-oyhr-juvgr plpyr naq riraghnyyl pnhtug. (I really like this one - there's even a deleted comment with a wrong answer I totally convinced myself was right and posted.) – Daniel Mehkeri Jun 26 '10 at 22:59
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    Is it impossible if the spiders are the same speed or slower? – Richard Dore Jul 08 '10 at 05:39
  • Lrf lbh pna. Cvpx nal fgengrtl. Qbrf abg znggre. Whfg znxr gur rzcgl cbfvgvba zbirf. Gur syl jbhyq nyjnlf zbir va na rzcgl cbfvgvba. Fb, va gvzr g, jura gur fcvqref pngpu gur svfu, gur ahzore bs rzcgl cbvagf ner g/(1+rcf). Naq gur ahzore bs cbfvgvbaf jurer gur syl jbhyq unir orra ner g. Gur pevgvpny pbaqvgvba vf jura g-[g/(1+rcf)]=1 => g=1+(1/rcf) – Pratik Poddar Nov 26 '10 at 12:40
25

Simplify (x-a)(x-b)...(x-z).

25

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?

Gjergji Zaimi
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Most of us know that, being deterministic, computers cannot generate true random numbers.

However, let's say you have a box which generates truly random binary numbers, but is biased: it's more likely to generate either a 1 or a 0, but you don't know the exact probabilities, or even which is more likely (both probabilities are > 0 and sum to 1, obviously)

Can you use this box to create a unbiased random generator of binary numbers?

BlueRaja
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  • Qb lbh jnag gb cynpr fbzr rkgen pbaqvgvbaf ba gur rssvpvrapl bs gur nytbevguz? Bgurejvfr: trg gjb ahzoref sebz gur obk. Vs lbh trg n bar sbyybjrq ol mreb, erghea bar. Vs lbh trg mreb sbyybjrq ol bar, erghea mreb. Bgurejvfr gel ntnva. Guvf grezvangrf nyzbfg fheryl, ohg pbhyq gnxr n juvyr... – Nate Eldredge Jun 25 '10 at 16:19
  • @Nate: Yep, that's the simplest solution I know of. Good dinner conversation puzzle because most people don't figure it out, but when you hear the solution it's seems so simple... – BlueRaja Jun 25 '10 at 18:49
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    I really like this one! – Andy Putman Jun 25 '10 at 20:18
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    Nate: Nal fbyhgvba pbhyq gnxr n juvyr. Vs gur obk vf irel ovnfrq, rnpu syvc whfg vfa'g cebqhpvat gung zhpu ragebcl. V nz abj phevbhf, ubjrire, vs gurer vf n fbzrjung zber rssvpvrag fbyhgvba. – Richard Dore Jun 25 '10 at 22:32
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    @Richard @Nate: Guvf cebprff vf nyfb pnyyrq haovnfvat be juvgravat (uggc://ra.jvxvcrqvn.bet/jvxv/Uneqjner_enaqbz_ahzore_trarengbe#Fbsgjner_juvgravat). Guvf cntr (uggc://jjj.pvcuretbgu.bet/pelcgb/haovnfvat/) pynvzf gb unir n fgengrtl juvpu vf arne vqrny, gubhtu V unira'g unq gvzr gb ernq vg lrg. – BlueRaja Jun 25 '10 at 22:40
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    znxr rnpu cnve bs ovgf bs gur byq trarengbe n ovg bs gur arj bar ol gbffvat bhg mreb-mreb naq bar-bar cnvef; bar pna nyfb erplpyr gur gbffrq bhg cnvef ol cnvevat gurz nf jryy (naq gbffvat mreb-mreb-mreb-mreb naq bar-bar-bar-bar); erplpyr naq ercrng... Vf guvf nf rssvpvrag nf vg pna trg? – Yaakov Baruch Jun 26 '10 at 20:18
  • @Yaakov: Lbh pna qb rira orggre ol gnxvat vagb nppbhag ubj zhpu cnvevat lbh ner qbvat... frr uggc://jjj.pbhefrf.snf.uneineq.rqh/~yvopf124/PF/pbvasyvc3.cqs – Armin Straub Aug 07 '10 at 05:15
  • Are the bits generated independent of each other? – Max Nov 29 '10 at 16:38
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    @RichardDore and others: the efficiency question leads to some interesting mathematics. See this paper by Yuval and my (slightly less, but still related) paper – Dan Romik Sep 13 '15 at 06:57
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A princess inhabits a flight of 17 rooms in a row. Each room has a door to the outside, and there is a door between adjacent rooms. The princess spends each day in a room that is adjacent to the room she was in the day before. One day a prince arrives from far away to woo for the princess. The guardian explains the habits of the princess and also the rules to him: Each day he may knock at an outside door of his choice. If the princess is behind it she will open and in the end marry him. If not, nothing happens, and he gets another chance the next day. Unfortunately his return ticket expires after 30 days. Does he have enough time to conquer the princess? (Adapted from "simpler-solutions.net")

  • Nice problem! The solution I found in the end is so simple that if I hadn't proved it, I wouldn't believe that it could work. :) – Nate Eldredge Jun 29 '10 at 19:36
  • I'm confused! It seems like an obvious solution is "ba qnl a, xabpx ba qbbe a." But if this works, why are there only 17 rooms? – Vectornaut Oct 13 '10 at 18:13
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    @Vectornaut The princess can move: she could spend day n in the room number (n+1). – Anthony Leverrier Oct 26 '10 at 15:42
  • Ba qnl a, xabpx ba qbbe a vs a vf yrff guna 17, be 33-a bgurejvfr. Rffragvnyyl gur nethzrag vf n qvfpergr irefvba bs fubjvat gung nal pbagvahbhf shapgvba sebz [0,1] gb [0,1] unf n svkrq cbvag. Bayl gung urer, gurl pna pebff rnpu bgure, ohg gur cevaprff'f cnevgl (bqq/rira) vf svkrq, juvyr jr fhccyl n "yvar" sbe obgu pnfrf. Jvyy guvf jbex? – BharatRam May 25 '11 at 17:16
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    This problem is addressed in general in Finding a princess in a palace: A pursuit-evasion problem (Britnell, John R. and Mark Wildon, 2012) and mentions this MO post specifically. Britnell and Wildon solve the problem for arbitrary graphs: they provide a strategy which is guaranteed to find the princess in bounded time if such a strategy exists, they characterize the graphs for which such a strategy does exist, and they show that their strategy provides the smallest bound among all such strategies. – Mark Dominus Oct 09 '19 at 19:12
  • @BharatRam 's solution doesn't work because rot13(lbh qba'g arrq gb xabpx ba qbbe bar). Actual solution: Purpx qbbef gjb gueh fvkgrra, gura fvkgrra gueh gjb. Guvf jbexf orpnhfr lbh fjnc cnevgl unysjnl gueh. – BlueRaja Dec 02 '19 at 14:48
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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?

Andrej Bauer
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    uneq sbe gur zngurzngvpvnaf, ohg ernyyl rnfl sbe gur tenaqzn... – Yaakov Baruch Jul 07 '10 at 21:57
  • V pna guvax bs zhygvcyr fbyhgvbaf - phg rdhny cvrprf sebz gur rqtrf naq fnir gur erfg sbe urefrys be sbe yngre; abg tvir gurz nal (gurl'ir unq rabhtu fjrrgf!); qvivqr vg vagb 8 rdhny fyvprf, naq chg vpvat ba n fyvpr bs gur svefg pnxr. Jung'f gur pbeerpg nafjre? – BlueRaja Jul 08 '10 at 21:02
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    On a related note, here's a similar puzzle: Grandma wants to cut the first cake into 8 equal slices, but only making three cuts. How can she do it? – BlueRaja Jul 08 '10 at 21:03
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    @ 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 09 '10 at 10:29
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    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 '10 at 17:58
  • I suppose it's out of the question to remove the icing from the sides, cut it into even cubes, and then redistribute the reserved icing fairly? – Ryan Reich Jul 27 '10 at 00:41
  • (I hope that's not a spoiler, it seems too absurd) – Ryan Reich Jul 27 '10 at 04:29
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    @Ryan: no, no, can't remove the icing of course. – Andrej Bauer Jul 27 '10 at 06:22
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    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 '10 at 23:27
18

You have a glass of red wine and a glass of white wine (of equal volume). You take a teaspoon of the red wine and put it in the glass of white wine and stir. You then take a teaspoon of the white wine (which now has a teaspoon of the red wine in it) and put it in the glass of red wine and stir.

Which glass has a higher ratio of (original wine)/(introduced wine)?

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    This is a classic so chances are everyone will know this at a dinner. – algori Jun 24 '10 at 04:59
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    You don't even have to stir perfectly – Ben Oct 27 '10 at 17:09
  • @Ben: Why don't you have to stir? First I was surprised. But the simplest solution is this: Take a very big teaspoon holding all red whine. –  Jun 29 '13 at 11:47
  • @Hilbert: Well, the solution works even without perfect stirring. – Ben Jun 29 '13 at 17:21
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    @Ben: O, what a fool I was! Consider simply every glass containing 100 marbles, red and white, at the beginning, and 100 marbles of both colours, at the end. Really no stirring necessary! –  Jun 30 '13 at 11:42
  • Right, it is remarkably easier to think about a finite set of marbles. – Ben Jul 01 '13 at 11:53
18

What is the resistance between 2 adjacent vertices of an infinite checkerboard if every edge is a 1 ohm resistor?

reb
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A certain rectangle can be covered by 25 coins of diameter 2. Can it always be covered with 100 coins of diameter 1?

Richard Dore
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16

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

ACL
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    Please define "left". (Eg.: the side with a wedding band...?) – Yaakov Baruch Jan 27 '11 at 04:05
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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 '11 at 10:02
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    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 Swett Mar 31 '12 at 14:36
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  1. Alice shuffles an ordinary deck of cards and turns the cards face up one at a time while Bob watches. At any point in this process before the last card is turned up, Bob can guess that the next card is red. Does Bob have a strategy that gives him a probability of success greater that .5?

  2. Let $x_1, x_2, \dots, x_n$ be $n$ points (in that order) on the circumference of a circle. Dana starts at the point $x_1$ and walks to one of the two neighboring points with probability $1/2$ for each. Dana continues to walk in this way, always moving from the present point to one of the two neighboring points with probability $1/2$ for each. Find the probability $p_i$ that the point $x_i$ is the last of the $n$ points to be visited for the first time. In other words, find the probability that when $x_i$ is visited for the first time, all the other points will have already been visited. For instance, $p_1=0$ (when $n>1$), since $x_1$ is the first of the $n$ points to be visited.

  3. Let $\pi$ be a random permutation of $1,2,\dots,n$ (from the uniform distribution). What is the probability that 1 and 2 are in the same cycle of $\pi$?

  4. Choose $n$ points at random (uniformly and independently) on the circumference of a circle. Find the probability $p_n$ that all the points lie on a semicircle. (For instance, $p_1 = p_2 = 1$.) More generally, fix $\theta<2\pi$ and find the probability that the $n$ points lie on an arc subtending an angle $\theta$ .

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    The answer to 3 surprised me, as did the brevity of the proof I found: Ercerfrag n crezhgngvba bs {1,...,a} ol, re, n crezhgngvba bs {1,...,a} nf sbyybjf: Jevgr qbja gur ryrzragf bs vgf plpyrf, rnpu plpyr ortvaavat jvgu vgf fznyyrfg ryrzrag, plpyrf va qrfpraqvat beqre bs gung fznyyrfg ryrzrag. Rirel crezhgngvba C bs {1,...,a} vf gur abgngvba sbe n havdhr crezhgngvba D bs {1,...,a}. Gur plpyr va D gung pbagnvaf 1 ortvaf jurerire 1 nccrnef va C naq pbagvahrf gb gur raq bs C; gurersber gung plpyr pbagnvaf 2 vss 1 cerprqrf 2 va C, naq jr'er qbar. – Gareth McCaughan Sep 30 '15 at 13:01
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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.)

Gil Kalai
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  • Nice problem. Here is a slightly different problem that can be solved in a similar way. Give these guys a strategy for guessing so that everybody guesses correctly with probability 1/100. – Tony Huynh Jul 09 '10 at 10:59
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    Very nice! Jbex zbq 100. Crefba a thrffrf (a zvahf (fhz bs ahzoref gurl frr)). Gura crefba (fhz bs nyy ahzoref) vf pbeerpg. – Peter LeFanu Lumsdaine Jul 12 '10 at 17:35
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Not a very difficult one but I like it since it is even suitable for non-mathematicians:

A small boat carrying a heavy stone is floating in a swimming pool. What happens to the level of water (up, down or remains equal) in the swimming pool if one removes the stone from the boat and throws it in the swimming pool?

The very easy solution suggests the following joke (illustrating the well-known ignorance of mathematicians of reality): Instead of sending scores of ships for saving passengers from the Titanic, one should have sunk all possible rescue-ships in order to lower the sea-level.

Roland Bacher
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    I feel I should point out that it's dependent on having a very dense (denser than water) stone rather than just heavy. Other than that, good question. – Mark Bell Jun 24 '10 at 14:03
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    If the stone sinks, it's denser than water. – Alison Miller Jun 26 '10 at 22:26
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    If you look on the periodic table, you'll find out that stone atoms are about twice as heavy as water atoms. As all atoms are the same size, and all solids and liquids pack their atoms as dense as possible, stone is necessarily twice as dense as water. – Theo Johnson-Freyd Jul 07 '10 at 22:43
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    Theo, your argument doesn't make much sense: "stone atoms"? Rocks contain many different elements, including hydrogen and oxygen (which water molecules contain). Also, solids don't necessarily pack atoms to maximize density (the heaviest solids aren't necessarily made of the heaviest atoms). Regardless, I think it's pretty safe to say stones tend to be heavier than water. Most kayakers don't worry about boulders floating down rivers and rolling over them. – Darsh Ranjan Aug 07 '10 at 03:10
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Instead of recommending some puzzles, I'll recommend some books containing many puzzles. Peter Winkler, Mathematical Puzzles; Peter Winkler, Mathematical Mind-Benders; Miodrag Petkovic, Famous Puzzles of Great Mathematicians.

Gerry Myerson
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12

Via the great Martin Gardner: A cylindrical hole is drilled straight through the center of a solid sphere. The length of hole in the sphere (i.e. of the remaining empty cylinder) is 6 units. What is the volume of the remaining solid object (i.e. sphere less hole)? Yes, there is enough information to solve this problem!

  • This is an amusing exercise for calculus courses (solids of revolution), although calculus is not necessary. – Michael Lugo Jun 24 '10 at 17:42
  • @Michael: Indeed! I first saw this as an exercise while I was teaching calculus and was somewhat surprised/annoyed by how long it took me to figure out. – Kiochi Jun 24 '10 at 18:59
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    It's easy to see what the answer has to be once you're told there's an answer, but I don't see how to prove it without setting up an integral. What am I missing? – Jonah Ostroff Jun 26 '10 at 14:54
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    Jonah, I don't think you're missing anything. To solve rigorously, I think you need calculus. Gardner's own answer basically said that, assuming you understand that the problem is well-posed, the answer must not depend upon the size of the sphere. So take the sphere with hole of diameter approaching zero (i.e. a sphere of diameter 6 units), which is just the whole sphere (i.e. 36π cubic units). –  Jun 27 '10 at 15:40
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    Full-blown calculus isn't necessary. You only need Cavalieri's principle: http://en.wikipedia.org/wiki/Cavalieri%27s_principle#The_napkin_ring_problem – aorq Jul 01 '10 at 18:12
10

(I learned this problem from Persi Diaconis.) A deck of $n$ different cards is shuffled and laid on the table by your left hand, face down. An identical deck of cards, independently shuffled, is laid at your right hand, also face down. You start turning up cards at the same rate with both hands, first the top card from both decks, then the next-to-top cards from both decks, and so on. What is the probability that you will simultaneously turn up identical cards from the two decks? What happens as $n \to \infty$? And does the answer for small $n$ (say, $n=7$) differ greatly from $n=52$?

10

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

Simplify:

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

Tom Boardman
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9

Countably many little dwarfs are going to their everyday work to the mine. They are marching and singing in a well-ordered line (by natural numbers), so that number 1 watches the backs of all the other ones, and, in general, number n watches the backs of all the others from n+1 on. Suddenly, an evil wizard appears on the top of a small hill, and magically puts a name on the back of each dwarf. Any name may be used, even more than once: existing ones, old-fashioned ones, or just weird sounds sprang out of his sick imagination, included grunts, sneezes and any snort-like name (you may enjoy providing your listerners with examples if they ask for). Then, he claims that, at his signal, everybody has to guess his own name, and say it loudly, all together. Whoever fails, will disappear immediately. Poor dwarfs are not new to these bully spells, and do have a strategy, that allows all but finitely many of them to survive. How do they do? To formalize, we may think the evil wizard has attached a real number to each dwarf's back.

Pietro Majer
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    I now see that this is a variant of Anton Geraschenko's problem. – Pietro Majer Jun 24 '10 at 05:47
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    Here is my solution: the strategy of the dwarfs is simply...n yvarne cebwrpgbe C sebz gur irpgbe fcnpr K bs nyy erny frdhraprf bagb gur fhofcnpr L bs nyy riraghnyyl inavfuvat erny frdhraprf. Gur jvmneq unf pubfra na ryrzrag k bs K; rnpu qjnes xabjf vg hc gb na ryrzrag bs L, fb ur pbzcyrgryl xabjf (V-C)k. Gurl nterr gb orunir nf vs Ck jrer 0, juvpu vf gur yhpxl pubvpr sbe nyy qjnesf ohg gur barf jvgu vaqrk va gur fhccbeg bs Ck. Bs pbhefr, gur rkvfgrapr bs gur cebwrpgbe vf rafherq ol gur nkvbz bs pubvpr. Jurer gurl qvq trg bar, vg erznvaf n zlfgrel. – Pietro Majer Jun 27 '10 at 14:58
9

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.

BlueRaja
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    Hint: lbh arrq gb ohea frira ebcrf – BlueRaja Aug 06 '10 at 17:58
  • Solution: Yvtug bar raq bs nyy 7 ebcrf, naq gur bccbfvgr raq bs bar bs gur ebcrf, ba sver fvzhygnarbhfyl. Gur bar jvgu obgu raqf yvg jvyy ohea bhg va rknpgyl 32 zvahgrf. Ng gur rknpg zbzrag vg oheaf bhg, yvtug gur bgure raq bs gur frpbaq bar. Vg jvyy ohea bhg va 16 zvahgrf. Jura vg svavfurf, yvtug gur guveq, juvpu jvyy ynfg 8 zvahgrf. rgp. – BlueRaja Jan 27 '15 at 23:00
8

You have 1000 bottles of wine. Exactly one of the bottles contains a deadly poison, but you don't know which one. The killing time of the poison varies from person to person, but death is imminent in at most $t$ hours after ingestion. You are allowed to use 10 notorious criminals as poison fodder (they are on death row). How much time do you need to correctly determine the poisoned bottle?

Tony Huynh
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  • How deadly is the poison? – Michael Lugo Jun 24 '10 at 17:44
  • It's fatal, but you don't know exactly how long it will take for each person. It's safe to assume that it will take roughly t hours for each person, but not exactly t hours. Otherwise you could determine the poison bottle in slightly over t hours if you had a precise enough stop watch. – Tony Huynh Jun 24 '10 at 19:14
  • To Tony Huynh: I think the question is not how long it takes for the drinker to die, but what amount of the drink is enough for that. – Zsbán Ambrus Jun 25 '10 at 22:07
  • Oh, I see. The poison is so deadly that even if someone drinks any amount (no matter how small) of the poisoned wine they will die in at most t hours. – Tony Huynh Jun 26 '10 at 16:07
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    Since the death penalty is occasionally dealt to innocent people, I would use 10 notorious politicians. – Yaakov Baruch Jun 27 '10 at 09:12
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    I think the first step is to dash to the store for 2 dozen bottles of wine. – Eric Tressler Jul 14 '10 at 15:29
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    There's a question about generalizing to k poisoned bottles at MU: http://math.stackexchange.com/questions/639/logic-problem-identifying-poisoned-wines-out-of-a-sample-minimizing-test-subjec/ – Qiaochu Yuan Jul 29 '10 at 17:50
7

There is a square with seven monkeys on the floor and seven bananas on the top. Seven ladders go up the square, from one monkey to the banana over it, and the monkeys can climb them. Moreover there are some ropes which connect the ladders.

A monkey will go up towards the bananas, but whenever it meets a rope it cannot resist the temptation to stray and hang on it. Prove that every monkey will reach a banana, no matter the configuration of ropes.

There are at least two different solutions to this.

Andrea Ferretti
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  • I guess we can assume that the ends of two ropes do not coincide? – Tom Smith Jun 24 '10 at 19:45
  • Yes, that is correct. Otherwise the path of a monkey would not be well defined. Maybe I should add, just to be precise, that there is a finite number of ropes. – Andrea Ferretti Jun 25 '10 at 09:44
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    Vaqhpgvba ba gur ahzore bs ebcrf: Jvgu n ebcr wbvavat cbvagf N naq O, lbh erzbir gur ebcr, phg gur ynqqref ng N naq O naq vagrepunatr gur gbcf, erfhygvat va na rdhvinyrag fvghngvba jvgu bar srjre ebcr. – Adam P. Goucher Dec 29 '13 at 13:32
7

When they came to diner some shook hands. Ask them to prove that that two of them shook hands the same number of times.

rgrig
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7

Oldie but a goodie (Monty-hall problem):

You are on a game show with three doors, behind one of which is a car and behind the other two are goats. You pick door #1. Monty, who knows what’s behind all three doors, reveals that behind door #2 is a goat. Before showing you what you won, Monty asks if you want to switch doors. Should you switch?

BlueRaja
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  • Jr nyy xabj gur nafjre vf, lrf, lbh fubhyq. Gur rnfvrfg jnl gb rkcynva vg gb aba-zngurzngvpvnaf vf: Jung vs gurer jrer 100 qbbef, jvgu 99 tbngf, naq nsgre cvpxvat bar qbbe, gur ubfg bcrarq hc 98 qbbef bs tbngf - jbhyq lbh fjvgpu gura? – BlueRaja Jun 25 '10 at 00:29
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    You should specify that Monty knows what's behind all three doors and is guaranteed to open a door hiding a goat (though I think this is intuitively understood). – JBL Jun 27 '10 at 01:46
  • After arriving at the solution, read this comic: http://spikedmath.com/020.html – Kiochi Jun 29 '10 at 15:04
  • If you need a car you should switch. Otherwise stay with door number 1. – Mykie Jul 28 '10 at 19:44
7

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?

7

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...

Dave Futer
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  • If 50 have one apple and 49 and one orange, then you can only select 25 bags with apples and 25 bags with oranges, which is not more than half the apples. Are you sure you didn't mean at least half the total number of apples/oranges? – BlueRaja Aug 06 '10 at 21:14
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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 strictly positive number of each fruit, then you can also get a strict inequality. – Dave Futer Aug 06 '10 at 21:17
6

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).

Pietro Majer
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6

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?

Gerald Edgar
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6

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.)

Ian Agol
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5

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.

Gjergji Zaimi
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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 05:55
5

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.

rgrig
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  • Also, Rustan Leino tends to entertain people with puzzles that end up on this list: http://research.microsoft.com/en-us/um/people/leino/puzzles.html – rgrig Jun 24 '10 at 16:01
5

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?

BlueRaja
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  • It seems to me there is something weird about "unknown process." What is a "process" for choosing numbers? Can there be uncountably many such processes? – Kiochi Jun 29 '10 at 05:07
  • @Koichi: You can think of "process" as any (of the uncountably-many) of $f(x): \mathbb{N} \to \mathbb{R}$, though this definition is not necessary to solve the problem. – BlueRaja Jun 29 '10 at 14:53
  • I'm calling shenanigans on this one ... it sounds like a flawed version of the wallet problem (which involves payoffs, http://en.wikipedia.org/wiki/Two-envelope_paradox#History_of_the_paradox). Do you know a strategy? – Daniel Mehkeri Jul 04 '10 at 18:53
  • @Daniel: That problem is unrelated. Guvf chmmyr jnf cbfrq ba Enaqnyy Zbaebr'f oybt (gur perngbe bs gur jropbzvp KXPQ). Gur ceboyrz, naq vgf fbyhgvba, pna or sbhaq urer: uggc://oybt.kxpq.pbz/2010/02/09/zngu-chmmyr/pbzzrag-cntr-1/#pbzzrag-17804 . Frr nyfb uggc://zngubiresybj.arg/dhrfgvbaf/9037/ – BlueRaja Jul 04 '10 at 20:35
  • Huh, well, borderline shenanigan, but I'd pay up, if we had money on it. :-) – Daniel Mehkeri Jul 05 '10 at 21:30
  • Deja vu: http://mathoverflow.net/questions/9037/how-is-it-that-you-can-guess-if-one-of-a-pair-of-random-numbers-is-larger-with-p/ – Darsh Ranjan Aug 07 '10 at 03:23
  • Has the case that one of the numbers is necessarily double the other been discussed anywhere here? This is how I've heard this problem:

    Take $x (a positive real amount), put $(2/3)x in envelope X, $(1/3)x in envelope Y, give each to each of Amy and Barnaby (with 50% liklihood of getting either) and tell them your process, i.e., that with 50% likilihood the other person got twice what you did, otherwise they got half what you did. Let them see their money if they wish, and ask them if they would like to switch. Both calculate an expected value of $1.25x for switching. Are they wrong?

    – AndrewLMarshall Dec 16 '10 at 21:51
5

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.

Gerhard "Ask Me About System Design" Paseman, 2010.07.28

5

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???

Answer: Read on from Here.

Anweshi
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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 06 '10 at 21:43
  • I wonder if this problem was inspired by Edgar Allan Poe's "The Gold-Bug," which contains a search for a treasure on an island using a tree and a specified direction. The search does not require any math (although the message telling them where to search is written in code, and the story explains how to solve a simple substitution cipher). I imagine a mathematically inclined reader liking the story, but wanting to improve how Poe's character hid the treasure. I bet Poe would have liked this problem. – anon Dec 08 '10 at 00:26
5

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?

doetoe
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  • Solution: Chg gur pbva rknpgyl va gur zvqqyr. Gura jurerire lbhe nqirefnel chgf gur pbva, chg vg ng gur fnzr cbfvgvba nsgre ebgngvba ol 180 qrterrf nebhaq gur pragre. (Guvf cbfvgvba vf nyjnlf serr). – doetoe Aug 09 '10 at 16:29
5

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?

4

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.

Kiochi
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4

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?

BlueRaja
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    Good thing SOMEONE solved this problem, otherwise we wouldn't have email! – Dylan Wilson Jun 25 '10 at 04:36
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    @Dylan: Actually, email is not normally encrypted or authenticated - you have to go through great pains (http://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
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    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
  • V'z abg fher guvf dhrfgvba vf cuenfrq irel jryy. Vg frrzf gb zr gur nafjre vf cebonoyl gung Wna fubhyq unir Znevn znvy uvz na haybpxrq obk (be na haybpxrq cnqybpx) fb gung ur pna gura frpheryl znvy ure onpx n evat va gung obk (be va n obk ybpxrq jvgu gung cnqybpx). Ohg jul qba'g gur guvrirf fgrny guvf cnqybpx/obk? V fgvyy yvxr guvf nafjre, naq V jvfu V pbhyq guvax bs n orggre jnl gb cuenfr gur dhrfgvba. Be znlor gurer'f n orggre nafjre? – Dan Ramras Jun 27 '10 at 16:29
  • @Dan: Nalguvat va na haybpxrq obk jvyy trg fgbyra, vapyhqvat cnqybpxf. Na rzcgl haybpxrq obk jba'g trg fgbyra, ohg ubj pna gung or hfrq gb fraq gur evat? – BlueRaja Jun 27 '10 at 16:33
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    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 01:33
  • @JBL: Rknpgyl evtug! Guvf vf eryngrq gb choyvp-xrl fpurzrf va juvpu rapelcgvba/fvtavat vf pbzzhgngvir (vr. R1(R2(z)) = R2(R1(z))), fhpu nf EFN. Guvf tvirf hf gur novyvgl gb unir fbzrbar fvta n qbphzrag jvgubhg orvat noyr gb ernq vg! Guvf vf vzcbegnag, sbe vafgnapr, sbe fbzr aba-erchqvngvba fpurzrf (uggc://ra.jvxvcrqvn.bet/jvxv/Aba-erchqvngvba), nf jryy nf fpurzrf juvpu erdhver gur trarengvba bs n zhghny enaqbz ahzore orgjrra gjb cnegvrf va juvpu arvgure cnegl "purng" (vr. pubbfr be ovnf gur ahzore). (pbag..) – BlueRaja Jun 28 '10 at 02:21
  • (pbag..) N fvzvyne fpurzr pna or hfrq gb perngr n flfgrz sbe pneq tnzrf jvgubhg n prageny freire va juvpu ab cynlre unf gur fnzr pneqf nf nal bgure, lrg ab cynlre pna xabj jung pneqf nal bgure cynlre unf. – BlueRaja Jun 28 '10 at 02:21
4

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.

Nate Eldredge
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Tony Huynh
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  • I suppose that each prince is able to observe the state of the switches only when he is called to the room (not from his cell), but before he chooses which switch to toggle? – Nate Eldredge Jun 29 '10 at 19:33
  • @Nate: That is correct. – Tony Huynh Jun 30 '10 at 02:19
  • This went around when I was in undergraduate, as a two-stage problem: posed first with the initial state of the switches known, then with it unknown. Among mathematician friends, we all got the first stage fairly quickly, but couldn't find the extension to the second case. My engineer housemate couldn't do the first part, and eventually asked for the solution; then immediately said “oh, you just need to add error-tolerance”, and extended it to the second case! – Peter LeFanu Lumsdaine Jul 12 '10 at 17:52
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    Actually, they can escape before they are old. Many generalizations of the problem here: http://www.segerman.org/prisoners.pdf – Dave Futer Aug 06 '10 at 22:22
4

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.

Ken Fan
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  • There is a particular balance scale that can be used at most 3 times, but the puzzle doesn't say we can't use a different scale! – Michael Burge Aug 15 '10 at 02:19
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    Sorry...you're not allowed to use more than one scale...you have to use a particular balance scale at most 3 times to solve the puzzle! I think in these puzzles it's also understood that all a balance scale can tell you is whether the stuff you put on one side is the same weight or not as the stuff you put on the other side, and if they aren't, which side is the heavier. – Ken Fan Aug 18 '10 at 17:32
  • I've asked about this on puzzling.SE because we couldn't figure out the answer: https://puzzling.stackexchange.com/questions/92254/are-all-balls-the-same-weight – BlueRaja Dec 26 '19 at 20:35
4

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?

Tracy Hall
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  • I learned this from Joe Buhler (over lunch, of course) who learned it from someone else. Hint: N fgengrtl jbexf cresrpgyl vs naq bayl vs vg jbexf bapr naq xrrcf jbexvat haqre nqwnprag genafcbfvgvbaf. – Tracy Hall Jul 30 '10 at 11:25
4

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?

jonderry
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  • Since the prisoner has infinite tries, he is guaranteed to win eventually by flipping random coins... unless the guard is allowed to be malicious (ie. he doesn't just turn the table randomly)? – BlueRaja Aug 06 '10 at 21:09
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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
  • @jonderry: very nice problem, I hadn’t come across it before! I’ve taken the liberty of editing it slightly to clarify BlueRaja’s misunderstanding: “spin” somewhat suggests randomness, so I’ve changed it to “turn”. Also: as I understand it, the coins/dice are spaced evenly around a circle on the turntable, rigth? i.e. at the vertices of a regular n -gon — is that right? – Peter LeFanu Lumsdaine Nov 29 '10 at 00:49
3

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.

JBL
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  • Stolen from http://concretenonsense.wordpress.com/2010/06/24/some-gems-from-physical-mathematics where it is written, "I learned of the third problem from Tadashi Tokieda’s (seems like a really fun guy, by the way) article “Mechanical Ideas in Geometry” in an old Monthly." Of course, that link contains a solution, so don't follow if you want to work on it yourself! – JBL Jun 26 '10 at 14:40
3

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.

Thierry Zell
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  • In fact your problem 3 is well-known; it was question A1 on the 2009 Putnam examination. Of course you may work with things other than the reals, but for dinner conversation this is not necessary!

    A fun generalization of this problem is as follows: for which of the following classes of quadrilaterals does the problem still hold: general quadrilaterals, parallelograms, rhombi, rectangles, trapezoids? (The trapezoids case is particularly devilish; I don't know the answer.)

    – dvitek Aug 11 '10 at 18:40
  • @drvitek: Thanks for your remark (especially since I couldn't remember which math competition I got 3 from). To even add to the origins of the problem and generalizations, in the solutions posted on the MAA website, they note:

    Problem 1 of the 1996 Romanian IMO team selection exam asks the same question with squares replaced by regular polygons of any (fixed) number of vertices.

    – Thierry Zell Aug 12 '10 at 12:03
3

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.

Edit. Actually, there is a strategy that works for 124 cards, but it is probably not human implementable (unlike the 52 card problem).

Tony Huynh
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  • Jryy whfg gur beqre jr fubj obo gur pneqf nybar tvirf hf 4! = 24 cbffvovyvgvrf. Fvapr gurer ner bayl 48 cbffvovyvgvrf sbe gur svsgu pneq, vs jr pbhyq guebj va nabgure snpgbe bs 2 fbzrubj (cerfhznoyl ol bhe pubvpr bs juvpu bs gur svir pneqf abg gb fubj Obo), jr'q unir n fbyhgvba. – BlueRaja Aug 10 '10 at 17:32
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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
2

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 08:32
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    I actually did say "five thousand" before reading a bit closer. – Michael Lugo Jun 26 '10 at 12:50
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    Qiaochu - If that's the case then it's not so much a puzzle as an attempted trick. – Mark Jun 27 '10 at 06:41
2

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.

schnitzi
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  • 1. Cvpx gur fznyyrfg cevzr > a, fvapr guvf vf gur fznyyrfg ahzore ≡ 0 zbq 2, zbq 3, ..., zbq a. 2. Nffhzvat rirelbar cvpxf enaqbzyl, cvpx gur ynetrfg cevzr c ≤ a, fvapr gur bayl jnl lbh jvyy trg cvpxrq vf vs lbh orybat gb n plpyr bs rknpgyl gung nzbhag. 3. Vs lbh hfr gur orfg fgengrtl (naq gurl xabj vg), frggvat hc n plpyr bs rknpgyl c jvyy trg lbh nal gvzr lbh cvpx nalbar va gung plpyr; naq vs (gurl xabj) lbh cvpx enaqbzyl, rirelbar cbvagvat ng lbh jvyy trg lbh sybbe(a/2) bhg bs (a-1) gvzrf (vr. 1/2 vs a vf bqq, bire 1/2 vs a vf rira) - qba'g xabj vs V pna cebir gung'f bcgvzny gubhtu. – BlueRaja Jul 19 '10 at 22:35
2

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?

villemoes
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  • It has already appeared on MO (though not in this thread): http://mathoverflow.net/questions/9754/magic-trick-based-on-deep-mathematics/28134#28134 – JBL Jul 29 '10 at 21:45
2

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.

  • Since this answer is incorrect, I assume it's okay to blow your rot-13 cover. I don't see any sense in which I x I contains lines. – Steven Landsburg Dec 07 '10 at 14:38
  • Yes, my previous comment was nonsense. and I've removed it. – Nate Eldredge Dec 07 '10 at 16:14
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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 08 '10 at 08:33
2

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.)

Florian
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1

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

BlueRaja
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1

Fork in the road 1

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.

What do you ask?

BlueRaja
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  • Frrzf gb zr gung vg'f fhssvpvrag gb nfx bar dhrfgvba gb bar angvir. "Vs, ulcbgurgvpnyyl, V jrer gb nfx lbh jurgure gur yrsg cngu yrnqf gb fnsrgl, jbhyq lbh nafjre lrf?" Nal angvir jvyy nafjre "lrf" vs gur yrsg cngu vf fnsr, naq "ab" vs vg vf abg. Hayrff gur ynathntr fcbxra ba guvf vfynaq ynpxf gur fhowhapgvir zbbq... – Nate Eldredge Jul 01 '10 at 15:27
  • @Nate: Yep, exactly :) – BlueRaja Jul 01 '10 at 16:53
  • Raymond Smullyan's "What is the name of this book?" has many more such puzzles. – Koundinya Vajjha Jul 12 '10 at 15:41
1

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.

user7361
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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 06 '10 at 05:45
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    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 06 '10 at 05:47
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    We've had this puzzle before on MO, haven't we? – Yemon Choi Jul 06 '10 at 06:01
  • @Gerry, That's a good suggestion. If I need to change it then I will. @Yemon, I don't know if it's been on MO before, but it's not previously in this thread. – user7361 Jul 06 '10 at 06:21
  • Don't know about on MO, but it has appeared lots of other places, e.g. http://blog.tanyakhovanova.com/?p=168 – JBL Jul 06 '10 at 12:41
  • @Yemon: yes, and a generalization too: http://mathoverflow.net/questions/10193/generalization-of-the-shakehands-condom-puzzle – Qiaochu Yuan Jul 06 '10 at 19:33
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    One should mention that the technique involved in the solution is not recommended as a safe practice by experts. See, e.g., http://tinyurl.com/352vhao (tinyurl'ed because the original url is a spoiler). – Nate Eldredge Jul 06 '10 at 20:05
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    Gerry's version is clean, but it has a trivial solution: do each surgery one-handed. :-) – Dave Futer Aug 06 '10 at 20:35
1

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).

1

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.

(Answer at http://7c6j.sl.pt)

  • I assume that to bring this up at a party, you memorized it first... – Yaakov Baruch Nov 29 '10 at 21:13
  • I'm reminded of something from Abbott and Costello: I'm 40, and my niece is 10, so she's one-fourth my age. When I'm 45, she'll be 15, one-third my age. When I'm 60, she'll be 30, one-half my age. So when will she be my age? – Gerry Myerson Nov 29 '10 at 23:01
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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 '10 at 12:28
1

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.)

Jakob Katz
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1

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?

  • 1
    Could be anything from epsilon to 100. I suspect the actual problem is just to prove that it's non-zero. – Gerry Myerson Jan 27 '11 at 03:40
  • Hmm...showing it's non-zero seems too trivial, but indeed the percentage can be arbitrary as long as it's greater than zero. Perhaps the birds are distributed according to some known distribution on the wire, and we wish to find the expected value of this proportion? – Daniel Litt Jan 27 '11 at 03:44
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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 '11 at 04:37
  • Using Ocam's razor I would assume that $n$ crows are independently uniformly distributed on $[0,1]$. – Christian Blatter Jan 27 '11 at 13:30
  • Sorry, I meant 'what is the probability that a crow is looking at a crow that is also looking at it' rather than 'percentage'! The fun thing about the answer was that it was independent of the number of crows. I'll try to dig up the original puzzle. – Jim Ward Jan 27 '11 at 18:58
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    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 01 '11 at 17:07
  • Rather than independent, I think what was meant is that there is an asymptotic value. – Douglas Zare Sep 25 '11 at 19:57
0

Fork in the road 2

You're once again at a fork in the road, and again, one path leads to safety, the other to doom.

There are three natives at the fork. One is from a village of truth-tellers, one from a village of liars, one from a village of random answerers. Of course you don't know which is which.

Moreover, the natives answer "pish" and "posh" for yes and no, but you don't know which means "yes" and which means "no."

You're allowed to ask only two yes-or-no questions, each question being directed at one native.

What do you ask?

BlueRaja
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  • BX, V fbyirq gur svefg bar (Nfx bar: "Ner lbh sebz gur fnzr ivyyntr?", vs ur fnlf "Lrf" gura gur bgure bar unf gb or gur gehgu-gryyre, vs ur fnlf "Ab" gur bgure bar vf gur yvne - urapr jr whfg nfx gur bgure bar jurer gb tb.) Ohg guvf bar V pna'g penpx - nal uvagf? – Harun Šiljak Jun 26 '10 at 06:15
  • @H. M. Gung vf abg gur fbyhgvba gb gur svefg bar. V znqr vg irel pyrne gung gurl !pbhyq! obgu or sebz gur fnzr ivyyntr, naq obgu ivyyntrf !pbhyq! or ivyyntrf bs yvnef/gehgugryyref, ohg arvgure ner arprffnevyl gehr. – BlueRaja Jun 26 '10 at 16:13
  • Bu, gur svefg irefvba bs gur evqqyr (orsber gur rqvgvat) gevpxrq zr - V nffhzrq bar ivyyntr vf gur ivyyntr bs gehgu-gryyref, naq gur bgure vf gur ivyyntr bs yvnef. – Harun Šiljak Jun 26 '10 at 16:26
  • Since every riddle of mine but this one has been solved, I'll post the answer here: Gur gevpx vf gb znxr fher gur -frpbaq- bar jr fcrnx gb vfa'g gur enaqbz nafjrere; gura jr pna hfr gur fnzr gevpx jr hfrq va 'Sbex va gur Ebnq 1' gb svaq gur pbeerpg cngu. Sbetrg nobhg gur 'cvfu/cbfu' sbe n frpbaq. Jr nfx gur svefg crefba "Jung jbhyq lbh fnl vs V nfxrq lbh vs crefba O vf gur enaqbz nafjrere?" Vs ur vf gur yvne be gehgu-gryyre, jr'yy trg gur gehgu bhg bs uvz naq or noyr gb nfx gur frpbaq dhrfgvba gb n aba-enaqbz nafjrere. (pbag..) – BlueRaja Jul 08 '10 at 21:37
  • (pbag..) Vs gur svefg crefba vf gur enaqbz nafjrere, vg qbrfa'g znggre jung ur fnlf fvapr jr pna svantyr gur gehgu bhg bs rvgure bs gur bgure gjb. Guvf tvirf hf bhe nafjre, rkprcg sbe gur 'cvfu/cbfu' ceboyrz. Gb trg nebhaq gung, jr nygre bhe dhrfgvba gb or "Vs V jrer gb nfx lbh vs crefba O vf gur enaqbz nafjrere, jbhyq lbh fnl 'cbfu?'" Jurgure cbfu zrnaf lrf be ab, obgu gur yvne naq gehgu-gryyre jvyy nafjre 'cbfu' jura O vf gur enaqbz nafjrere, naq 'cvfu' jura ur vfa'g. Jr nfx fvzvyneyl sbe gur pbeerpg ebnq, naq jr ner qbar. – BlueRaja Jul 08 '10 at 21:39
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Bob and Alice want to marry each other, so Bob decides to send Alice a ring. The problem is that they both live in different countries, and any valuables they send through the mail are sure to be stolen, unless they are sent in a locked box. The box can be locked by a padlock which can only be opened by the right key. Both Alice and Bob have an infinite supply of boxes and padlocks with corresponding keys. However, neither Alice nor Bob have keys to each other's padlocks, only for their own. Suppose you can put boxes inside each other. How can Bob send Alice the ring? Of course, the solution to the problem must end with Alice putting the ring on her finger. To reiterate, anything outside of a padlocked box is guaranteed to be stolen.

This problem has numerous solutions as well as interpretations which makes for a fun discussion. It can also be solved in your head without pencil or paper.

Alex R.
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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

-1

A room contains 3 bulbs and 3 switches outside controlling the bulbs. Is it possible to determine which switch controls which bulb by entering the room only once and observing bulbs?

Bugs Bunny
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    guvf vf abg zngu! – Yaakov Baruch Jun 24 '10 at 12:15
  • yby, ebg13 vf abg qrnq :) – Harun Šiljak Jun 24 '10 at 14:10
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    http://en.wikipedia.org/wiki/ROT13 – Qiaochu Yuan Jun 24 '10 at 14:29
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    It should be noted that this has several possible answers depending on what kind of hardware is used in the question. – Ketil Tveiten Jun 24 '10 at 14:33
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    Actually you can solve it with $2^2$ bulbs, and only once entering the room. – Pietro Majer Jun 24 '10 at 18:06
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    The solution I know is non-mathematical and, in principle, allows you to solve the problem with an arbitrary number of bulbs. Do people have in mind a mathematical solution? – Qiaochu Yuan Jun 24 '10 at 22:28
  • I believe I have seen a mathematical (ie. not hardware-dependent) solution to this, but I can't remember it. – Ketil Tveiten Jun 25 '10 at 12:02
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    I feel like I used to know the answer, too. This is maddening.

    By the way, Bugs: Yesterday I read on a bumper sticker "Legalize Updoc".

    – Tom Goodwillie Jun 26 '10 at 04:21
  • @Ketil,Tom: Gurer ner fvk cbffvovyvgvrf sbe gur zngpuvat bs fjvgpurf naq yvtugohyof. Ohg vs lbh ghea bar fjvgpu ba, gurer ner bayl guerr cbffvovyvgvrf sbe jung lbh frr va gur ebbz. Fnzr vs lbh ghea gjb fjvgpurf ba. Fb yvtug vf vafhssvpvrag. Hfvat obgu urng naq yvtug jbexf, bs pbhefr. (How am I being non-mathematical?) – Daniel Mehkeri Jun 26 '10 at 16:47
  • fznpxf bja sberurnq – Tom Goodwillie Jun 26 '10 at 22:18
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    @Daniel: Sbe zr gur hfr bs urng vf "aba-zngurzngvpny" va gur frafr gung vg gnxrf cynpr bhgfvqr bs na vzcyvrq zngurzngvpny zbqry bs gur flfgrz, r.t. bar juvpu vf zrzbelyrff. – Qiaochu Yuan Jun 26 '10 at 22:20
  • @Qiaochu: Gura jul jbaqre nobhg jurgure crbcyr unq n zngurzngvpny fbyhgvba va zvaq vs lbh xarj gurer jnf abar? Vg fbhaqf yvxr lbh qvqa'g frr gur vzcbffvovyvgl cebbs. Fheryl gung cneg bs gur nafjre jnf ragveryl zngurzngvpny. Gurersber V pbapyhqr vg jnf n irel tbbq ceboyrz :-) – Daniel Mehkeri Jun 27 '10 at 03:40
  • V snvy gb frr ubj urng vf yrff zngurzngvpny guna yvtug. Vafgrnq bs cerfhzvat n zngurzngvpny zbqry, bar arrqf gb nanylfr gur erny yvsr fvghngvba naq pbzrf hc jvgu n orggre zbqry. – Bugs Bunny Jun 28 '10 at 10:43
  • Gur 'aba-zngurzngvpny' vffhr urer vf gur qrcraqrapr hcba uneqjner, r.t. gur nafjre punatrf vs lbh hfr YRQ ohyof vafgrnq bs svynzrag ohyof. – Ketil Tveiten Jul 25 '10 at 13:16
  • Regardless of hardware, I'd be interested in @Pietro's solution to solving it with four bulbs (assuming it's not just qvssrerapr va urng orgjrra gur guerr 'bss' ohyof, juvpu jbhyq nyybj hf gurbergvpnyyl hayvzvgrq ahzore bs ohyof...) – BlueRaja Aug 06 '10 at 21:19
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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!

  • The point of having a list is that people should read the answers before adding their two cents... – Yemon Choi Feb 02 '11 at 02:52
  • yes, I agree... and actually what I mean is that I didn't solve all the puzzles, and I don't know if some or them are equivalent to this one here. Can someone point me at equivalent puzzles? I don't think that the infinite cases are, because for the solution of this one you need finiteness (at least, for the solution I found!) – klaraspina Feb 02 '11 at 11:26