Authors: Belov D:: Bogdanov I.I., Knop K.A.

Initially, there are three piles of 100, 101, and 102 stones on the table, respectively. Ilya and Kostya are playing the following game. In one move, each of them can take one stone from any pile, except the one from which they took a stone on their previous move (on their first move, 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?
Let's outline a winning strategy for Ilya. Denote by $A, B$, and $C$ the piles that initially contain 100, 101, and 102 stones, respectively. Ilya's first move is to take a stone from pile $B$. There are two possible cases.

1) If Kostya's first move is to take a stone from a pile other than $B$, 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. This will leave an odd number of stones in $X$, 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.
2) If Kostya's first move is to take a stone from pile $B$, Ilya will follow this strategy: If Kostya takes a stone from pile $C$ at some point, Ilya will also take a stone from $C$. If Kostya takes a stone from pile $A$ or $B$, 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.

Let's show that Ilya will always be able to make a move in this case as well. Ilya will be able to make his second move (by taking a stone from $A$). If, at some step, Ilya needs to take a stone from pile $C$, then Kostya must have taken a stone from $C$ on the previous move, meaning that neither player took a stone from $C$ on the previous step. Since there is an odd number of stones in $C$ after Kostya's move, Ilya can take a stone from $C$. If Ilya needs to take a stone from a pile other than $C$, say from $A$, then Kostya just took a stone from $B$ (and there is still a stone in $A$). Therefore, 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.

## Answer

Ilya.

Submit a comment