Let $M$.

On the table, there are three piles of matches. The first pile contains 100 matches, the second contains 200, and the third contains 300. Two players play a game. They take turns, and on each turn, a player must remove one of the piles and divide any of the remaining piles into two non-empty parts. The player who cannot make a move loses. Who will win with correct play: the first player or their partner?
Let before some move, the number of matches on the table are in the form $2^{n} a, 2^{n} b$ and $2^{m} c$, where $0 \leq n < m$. If the starting player makes a move, they can always leave the number of matches in the form $2^{k} x, 2^{k} y, 2^{l} z$ where $k \leq l$ and $k \leq n$. If $l > n$, then $l=n$; the case $k=n$ is impossible. Thus, the starting player will again find themselves in a good position.

Since the initial position is good: $100=2^{2 \cdot 25}, 300=2^{2 \cdot} \cdot 75,200=2^{3 \cdot 25}$, then with this strategy, the starting player will always end up in a good position, meaning they will always have a move.

## Answer

The starting player.