This game will be familiar to many mathematicians, and it is always good fun to play.

I am looking to find a list of good questions with short, when-you-see-it solutions. The kind of question one could solve pretty much anywhere: over dinner, while taking a walk - essentially, without pen & paper, and where the solution lies in spotting a clever trick or fact (rather than via some monstrously power theorem).

To get an idea of what I mean, here is an example:

Q. A real number is called repetitive if its decimal expansion contains arbitrarily long blocks which are the same. Prove that the square of a repetitive number is repetitive.

(please do not post a solution, since I am sure anyone can figure it out given enough time)

Does anyone have similar chestnuts to offer?

  • 1
    $\begingroup$ See this, where Prof. Gowers asks pretty much the same question. Though I have to admit, I have never come across a problem with such a massive "aha" moment as that example sheet question. $\endgroup$ – Joshua Pepper Mar 24 '14 at 1:01
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    $\begingroup$ I still haven't solved the example which makes me kind of sad... $\endgroup$ – recursive recursion Mar 24 '14 at 1:07


$$\int_0^1\!\!\int_0^1\! \dfrac{1}{1-xy} \, \mathrm{d}x\mathrm{d}y$$

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    $\begingroup$ Good one. It took me a while. $\endgroup$ – recursive recursion Mar 24 '14 at 1:09

Consider a 8x8 chessboard and 2x1 dominos. Clearly one can easily cover the whole chessboard with 32 dominos.

Now omit the squares on the bottom-left and top-right. There are 62 squares left on the chessboard. Can you cover them using 31 dominos?


$1$, $z$, $z^2$, and $z^3$ (all distinct) lie on the same circle on the complex plane. What is the center of that circle?

  • $\begingroup$ with no calculus at all ? $\endgroup$ – Thomas Mar 26 '14 at 14:45
  • $\begingroup$ Well, you need to know something about complex numbers. What do you mean by calculus? $\endgroup$ – JiK Mar 27 '14 at 11:54
  • $\begingroup$ just wanted to be sure that you don't need a paper and a pen to solve this one $\endgroup$ – Thomas Mar 27 '14 at 16:52
  • $\begingroup$ The idea I came up with was to multiply all these points by $z$ and conclude that $|z|=1$. Is this the idea, or is there and even simpler way? $\endgroup$ – rschwieb Apr 10 '14 at 11:39
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    $\begingroup$ That seems pretty simple! Three points determine a circle, so $1, z, z^2$ determine the circle - which must therefore be invariant under multiplication by $z$, since $z, z^2, z^3$ determine the same circle. If $C$ is the center then this implies $C=Cz$ which immediately gives $C=0$, and the fact that $1$ is on the circle gives that it's the unit circle. $\endgroup$ – Steven Stadnicki Apr 16 '14 at 3:22

A rectangular room in the museum has a floor tiled with tiles of two shapes: $1 \times 4$ and $2 \times 2$. The tiles completely cover the floor of the room, and no tile has been damaged, cut in half, or otherwise had its integrity interfered with. One day, a heavy object is dropped on the floor and one of the tiles is cracked. The museum handyman removes the damaged tile and goes to the storage closet to get a replacement. But he finds that there is only one spare tile, and it is of the other shape. Can he rearrange the remaining tiles in the room in such a way that the spare tile can be used to fill the hole?

  • $\begingroup$ Nifty! Quick (hopefully not too spoilery) check: does the answer (or at least an answer) use a 'bricklayer' style coloring? $\endgroup$ – Steven Stadnicki Apr 16 '14 at 3:26
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    $\begingroup$ @Steven, that is one way to do it. You might also consider this diagram: junk.orderofthehammer.com/j2014/floor-tiling-puzzle-hint.png $\endgroup$ – Hammerite Apr 16 '14 at 13:35
  • $\begingroup$ @Steven Stadnicki what do you mean by 'bricklayer' style colouring? Can you please give a link, if possible? Thanks $\endgroup$ – user144361 Mar 31 at 7:14
  • $\begingroup$ @Hammerite: what is Bricklayer style colouring? Thank you. $\endgroup$ – user144361 Mar 31 at 7:32
  • $\begingroup$ Imagine a colouring that resembles a brick wall made up of 2x1 bricks. Each brick is offset by 1 from bricks below it. Bricks lying along diagonal lines are painted the same colour, giving a stripe effect. $\endgroup$ – Hammerite Mar 31 at 15:08

This is an old putnam problem, that I think is an excellent example of this.

Let $a_n = 1111...1 $($n$ ones) for instance $a_2 = 11$. Find all polynomials $f$ such that for all $a_n$, there is some $a_k$ where $f(a_n) = a_k$.

Hint: Reduce the problem to finding polynomials $g$ where $g(10^n)=10^k$.


If the probability of rain on Saturday is 50% and the probability of rain on Sunday is 50%, which is the probability of rain on the weekend? It can be very fun during dinner with non-mathematicians friends...

  • $\begingroup$ If rain on Saturday and rain on Sunday are independent, it's 75%. But they're not, which is probably where people's intuition comes in. As an engineer I'd say the answer is unknown but lies between 50% and 75%. If you'd asked about the 1st of the month and the 31st of the same month, I'd say 75% Alternatively, If there's 50% chance of rain at 11.59pm on Saturday and 50% chance of rain at 12.00am on Sunday, I'd say the chance of rain at 11.59 or 12.00 is pretty close to 50% as they're strongly dependent. $\endgroup$ – Level River St Jan 12 '18 at 1:28
  • $\begingroup$ @LevelRiverSt Between 50% and 100%, actually. What if the whether was such that after every non-rainy day, there's a rainy day? Then you could only have rainy days in succession but not non-rainy days. (Basically, this is how Parisian weather works in winter.) $\endgroup$ – yo' Jan 12 '18 at 12:52

Someone plays chess at least once a day and no more than 10 times in 7 consecutive days. Prove that there is a period of n consecutive days where the man or woman played 23 times.


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