I'm holding a 3-5 minute speech next week on mathematical problem solving, and how it makes me happy, to 15-20 non-mathematicians. As a part of it, I had thought about demonstrating two problems, but I can only come up with one. That would be the following:

I shuffle a deck of cards and deal each person one. Then without looking at it, they stick it to their forehead so that everyone else can see it. Now I will ask one of the players to say "red" or "black", at which point everyone else should know whether they have a red or black card, if they had had a strategy planned out.

Now, I'm looking for a second problem. These are the things I'm looking for: 1) An easily formulated problem with no "hard core" math involved. 2) An easily explained, but not too obvious, solution. 3) Something everyone can join in on. 4) Not too much preparation needed. 5) Something not related to the problem I already have, as in reasoning on what you know and can see, and you know other people can see and so on.

  • $\begingroup$ math.stackexchange.com/q/62486/15660 $\endgroup$ – Peđa Terzić Oct 28 '11 at 6:23
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    $\begingroup$ 3 to 5 minutes??? You really don't need two problems. Or did you mean a 35-minute speech? $\endgroup$ – Robert Israel Oct 28 '11 at 6:35
  • $\begingroup$ @RobertIsrael I guess you're right... This would take time. Guess I'll go with just the one. $\endgroup$ – Arthur Oct 28 '11 at 7:44

I find that people are really surprised by the Blue-Eyed Islander problem.

In brief summary, the inhabitants of an island have a policy that, if one learns ones own eye color, that person must commit suicide at dawn the next day. A foreigner visits and says, to no one in particular, "At least one person here has blue eyes."

What effect should this have on the islanders? Your audience will undoubtedly agree that nothing should happen - all the islanders can look around and see that there are people with blue eyes. The reality, however, is that all the blue-eyed people will commit suicide after a certain number of days.

  • $\begingroup$ You should add that he must commit suicide at the dawn of the next day, otherwise it won't work I guess. $\endgroup$ – Listing Oct 28 '11 at 7:38
  • $\begingroup$ @Listing You are correct, but I just wanted to briefly convey the flavor of the problem. Certainly, OP needs to read the problem in full by following the link before attempting to present it. $\endgroup$ – Austin Mohr Oct 28 '11 at 7:44
  • $\begingroup$ This is quite close to the problem that I already have. I might actually swap. Thanks for the input. $\endgroup$ – Arthur Oct 28 '11 at 7:44
  • $\begingroup$ @Arthur If you end up using it, I suggest you explain the cases of 1, 2, and 3 blue-eyed people in full detail before appealing to induction. My audiences always seem to think there is still something special going on for 2 blue-eyed people. Explicitly covering the case of 3 blue-eyed people generally satisfies them that the induction carries through. $\endgroup$ – Austin Mohr Oct 28 '11 at 7:48
  • $\begingroup$ One similar in the problem-solving method (deducing step-by-step) would be en.wikipedia.org/wiki/Pirate_game. It's not as similar as you might think, and it's quite fun. Who doesn't like pirates and gold? $\endgroup$ – Dustin Tran Oct 28 '11 at 7:56

A mixed bag of questions that I’ve used for various vaguely similar purposes in the past:

(1) A suitably dressed-up version of the result that every graph must have an even number of odd vertices.

(2) There is a heap of $1001$ stones on the table. Repeat the following operation: choose some heap containing more than two stones, throw one stone away from that heap, and divide the remaining stones into two (not necessarily equal) heaps. Is it possible to end up with only heaps of size three?

(3) Let $n$ be an integer greater than $1$. The integers $1, 2, 3, \dots, n^2$ are placed on the squares of an $n\times n$ chessboard, one integer per square. Show that no matter how this is done, there must be two adjacent squares whose numbers differ by more than $n$. (The squares may be adjacent horizontally, vertically, or diagonally.)

(4) Brynjulf and Kjellaug have three pieces of paper. At any time Brynjulf is allowed to pick up one piece of paper and tear it into three smaller pieces, and at any time Kjellaug is allowed to pick up one piece of paper and tear it into five smaller pieces; each of these operations is called a play. They are not allowed to play simultaneously, but they are also not required to take turns: Brynjulf might make three plays in a row, then Kjellaug might make two, Brynjulf one, Kjellaug $17$, and so on. Can they manage to finish a play with exactly $100$ pieces of paper?

(5) Six bowls are arranged in a circle on the table, with one marble in each bowl; this is the starting position of a solitaire game. The rules are simple: at each turn you must pick up one marble in each hand and move it to one of the two bowls immediately adjacent to the one from which you got it. You are not required to draw the marbles from different bowls: if some bowl contains more than one marble, you may take both from that bowl. If you do take two marbles from the same bowl, you need not move them to the same bowl: you may put one into each of the two neighboring bowls. The object of the game is to get all six marbles into a single bowl. Either explain how this can be done, or prove that it cannot be done.

(6) Variations on a theme: (a) Five darts are thrown at a square measuring $14$ inches on a side. Prove that two of them must be no more than $10$ inches apart. (b) Five darts are thrown at an equilateral triangle measuring $14$ inches on a side. Prove that two of them must be no more than $7$ inches apart. (c) Four darts are thrown at an equilateral triangle measuring $14$ inches on a side. Prove that two of them must be no more than $9$ inches apart. Can this maximum distance be improved to $8$ inches? (d) Nineteen darts are thrown at a regular hexagon measuring $12$ inches on a side. Prove that two of them must be no more than $7$ inches apart.

(7) Several people sit around a lunch table. As it happens, each person’s age is the average of the ages of the two people immediately to his or her right and left. Jessica says that she’s 26; Ian says that he’s only 24. How do I know that one of them is lying?


Suppose you have domino pices that each one can cover 2 squares chessboard. Take a square from bottom left and top right of the board. Can the chess board be covered with domino pices?


My personal favourites, which not all fit your requirements, are

  1. Cantor enumeration to show that $$ \mbox{card}(\mathbb{N}) = \mbox{card}(\mathbb{N^2}) $$ and other noteworthy facts about same and different infinities
  2. The Monty Hall problem
  3. The integration $$ I = \int\limits_{-\infty}^{\infty}e^{-t^2}dt $$ which needs the out-of-the-box thinking from Poisson that there is more than two ways to sum up over the plane.

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