10.3. Initially, there are three piles of 100, 101, and 102 stones on the table, respectively. Ilya and Kostya are playing the following game. On each turn, each of them can take one stone from any pile except the one from which they took a stone on their previous turn (on their first turn, each player can take a stone from any pile). The players take turns, with Ilya starting. The player who cannot make a move loses. Who among the players can win, regardless of how the opponent plays?

(D. Belov, I. Bogdanov, K. Knop)
Answer. Ilya.

Solution. We will show how Ilya can move to guarantee a win. Let $A, B$, and $C$ denote the piles that initially contain 100, 101, and 102 stones, respectively. Ilya's first move will be to take a stone from pile $B$. There are two possible cases.

Case 1. Kostya takes a stone from a pile other than $B$ on his first move. Then Ilya will always repeat Kostya's move, that is, take a stone from the same pile from which Kostya just took a stone. Note that, while Ilya follows this strategy, after each of his moves, there will be an even number of stones in each pile.

Ilya will be able to make his second move because Kostya took his first stone from a pile other than $B$. After this, each move by Kostya will involve taking a stone from a pile $X$ different from the pile $Y$ from which Ilya just took a stone. Then in $X$ there will be an odd number of stones, and Ilya will be able to take another stone from $X$. Therefore, by following this strategy, Ilya will always be able to make a move after Kostya's move. Since the game will eventually end, Kostya will lose.

Case 2. Kostya takes a stone from pile $B$ on his first move. In this case, Ilya will follow the following strategy. If before Ilya's current move, Kostya took a stone from pile $C$, then Ilya will take a stone from there as well. If Kostya took a stone from pile $A$ or $B$, then Ilya will take a stone from pile $B$ or $A$ respectively. Note that, while Ilya follows this strategy, after each of his moves, there will be an even number of stones in $C$, and the number of stones in $A$ and $B$ will be equal.

Again, we need to show that Ilya will always be able to make a move according to this strategy. Ilya will be able to make his second move (by taking a stone from $A$). If on a certain step he needs to take a stone from pile $C$, then Kostya must have taken a stone from there on the previous step, meaning that on the previous move, neither player took anything from $C$. At this point, after Kostya's move, there is an odd number of stones in $C$, so Ilya can take a stone from $C$. If Ilya needs to take a stone from pile $A$ or $B$, say from $A$, then Kostya just took a stone from $B$ (and there is still a stone in $A$). This means that on the previous step, Kostya could not have taken a stone from $B$, so Ilya did not take a stone from $A$, meaning he can take a stone from there on the current step.

Remark. The strategy described in Case 2 also works if Kostya takes a stone from $C$ on his first move.