This post contains a proposed solution to Problem #032 - restaurant roundtable. Please do not read this solution before making a serious attempt at the problem.

I am excited to tell you that I just released the alpha version of my “Pydont's” book, a book that compiles all my “Pydon't” articles. You can get the book at leanpub:


Congratulations to the ones that solved this problem correctly and, in particular, to the ones that submitted correct solutions:

  • Attila K., Hungary;
  • Filippo M., Italy;
  • Luís S.;
  • André S..

(The list is in no particular order.)


I received some really interesting solutions and I'll be sharing a paraphrasing of the one I think is the most elegant.

Yes, it is always possible to turn the table some number of places in order to fix the dishes of at least two people at the same time. Let's see how to do it.

Each person sitting at the table will look to its left and count the number of plates until its own plate.

For example, if we consider this placement:

Then the corresponding numbers would be the following:

If there are \(n\) people sitting at the table, that number is an integer between \(1\) and \(n - 1\), inclusive. It is not \(0\) because that would mean that person got the correct plate and it is not \(n\) because that is a whole turn around the table, also meaning that person would've gotten its own plate.

We have seen that each person has a number inside the set

\[ \{ 1, 2, \cdots, n-1 \} ~~~,\]

and there are \(n\) people, but that set only contains \(n - 1\) distinct numbers. Therefore, the pigeonhole principle tells you that at least two persons have the same number \(d\), which means you can turn the table \(d\) times in the counter-clockwise direction to deliver the correct plate to them.

If you have any questions about my solution, found an error (woops!) or want to share your solution, please leave a comment below! Otherwise just leave an “upvote” reaction!-->

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