10.2. Vasya and Petya are playing "Take or Split." In this game, there is initially one large pile of stones. On each turn, the player whose turn it is either splits one of the existing piles into two smaller piles in any way, or takes one of the existing piles. The player who leaves no stones on the field after their turn wins. The players take turns, with Vasya going first, but on his first and only first turn, he is not allowed to take the entire pile at once. Who will win this game? Note: a pile can contain just one stone.
Answer. Vasya will always win.

Solution. Let by the time the game ends, both players have made $\boldsymbol{x}$ moves of the type taking a pile and $\boldsymbol{y}$ moves of the type dividing a pile into two. Each move of the first type reduces the number of piles on the field by one, and each move of the second type increases this number by one. Initially, there was one pile, and at the end, there are zero, so we have the relation $1+y-x=0$, from which $x=y+1$. The total number of moves made in this case is $x+y=2 y+1$ - an odd number, therefore, the last move was made by the first player, that is, Vasya, and he won. The answer does not depend on the number of stones at the beginning.

Evaluation Criteria. ($\cdot$) Based on the consideration of particular cases, the correct answer is claimed: 1 point. ($\cdot$) The correct answer is proven for some infinite series of values of the number of stones in the pile: 2 points (not cumulative with the previous one). ($\cdot$) The idea of considering the sum $x+y$ is present: 1 point.