2. Given natural numbers $m$ and $n$. On the table, there are two piles: the first pile has $n$ stones, and the second pile has $m$ stones. Petya and Vasya play the following game. Petya starts. In one move, one can split one of the existing piles on the table into several smaller piles. The player who cannot make a move loses. For which $m$ and $n$ can Petya ensure his victory regardless of Vasya's play?

(A. Khryabrov)
Answer: $m \neq n$.

Solution. If $m=n$, Vasya can repeat Petya's moves and, therefore, will always be able to make a move. Let $m>n$. Then Petya will break the pile with $m$ stones into a pile with $n$ stones and $m-n$ piles of one stone each. Nothing more can be done with the piles of one stone, so the game will continue with two piles of $n$ stones. Here, Petya will repeat Vasya's moves and, therefore, will always be able to make a move.