14S59 ** There are two piles of matches on the table, with 100 and 252 matches respectively. Two people take turns to take matches, each time only from one pile, and the number of matches taken must be a divisor of the number of matches in that pile. The one who takes the last match wins. If both players adopt the correct strategy, who will win?
Parse "The first player to take must lose."
For each natural number $n$, let $v(n)=\max \left\{k\left|k \in \mathbf{N}, 2^{k}\right| n\right\}$. If there are $n$ and $m$ matches in two piles, and $v(n)=v(m)$, then this equality will certainly be disrupted after one move. The reason is as follows: Suppose $r$ matches are taken from the first pile. Since $r \mid \dot{m}$, it follows that $v(r) \leqslant v(m)=v(n)$. If $v(r) < v(m)$. In summary, the equality $v(n)=v(m)$ is indeed disrupted. If after the first player's move, there are $n$ and $m$ matches left in the two piles, and $v(n)>v(m)$, the second player can take $s=2^{v(m)}$ matches from the first pile, making the equality $v(n-s)=v(m)$ hold. Initially, $v(100)=v(252)=2$, so the second player can always maintain the equality $v(n)=v(m)$ after each move. Therefore, he must win.