this has been a riddle to difficult for me to understand the solution some year ago. Even though the riddle seems really easy.
You have got 55 coins. Someone builds an arbitrary number m of stacks of arbitrary size out of them. Each step you take away one coin from each stack and out of these you build a new stack with m coins. Repeating this process you eventually get ten stacks with size 1, 2, 3, 4, 5, 6, 7, 8, 9 and 10 (adds up to 55). Now the procedure will reproduce this constellation.
Question: Why does this constellation always occur?
Equivalent to: Why is there no other constellation that reproduces itself alfter a certain amount of steps? For example suppose it had been 4 coins. then $ (1,3) \rightarrow (2,2) \rightarrow (1,1,2) \rightarrow (1,3) $ is such a multi-step loop. If, there would be any multistep loop with 55 coins, we wouldn't reach the position $ (1,2,3,4,5,6,7,8,9,10) $, hence there can not be a multi-step loop. On the other hand, if there is no, then after finitely many steps we reach the $ (1,2,...,10) $ position, in order to not repeat ourself in any other way, which would be contradicting the non-existence of multi-step-loops. (There is no other single-step-loop than the $ (1,2,3,4,5,6,7,8,9,19) $ one. The proof is straight forward.)
You can translate this problem into Young Diagrams. The solution I overheard some working mathematicians discuss did this, but I was not able to follow the arguments back than. They themself claimed the solution to be quite hard. This was the final riddle of some big mathematicians-riddle contest, they said. I hope this is not a duplicate, I have searched the side and the web, but I didn't find anything.