# 6. CONDITION

There are 2001 coins on the table. Two players play the following game. They take turns. On a turn, the first player can take any odd number of coins from 1 to 99, and the second player can take any even number of coins from 2 to 100. The player who cannot make a move loses. Who wins with correct play? Justify your answer.
Solution. Let's describe the first player's strategy. The first move should be to take 81 coins from the table. For each subsequent move, if the second player takes $x$ coins, the first player should take $101-x$ coins. He can always do this because if $x$ is an even number from 2 to 100, then ($101-x$) is an odd number from 1 to 99. Since $2001=101 \cdot 19+81+1$, after 19 such "responses," there will be 1 coin left on the table after the first player's move, and the second player will not be able to make a move, i.e., he will lose.

Answer. The first player wins.