$8 \cdot 20$ There are two piles of matches on the table, with the number of matches being 100 and 252, respectively. Two people, A and B, take turns to pick up matches. Each time, they can only take matches from one pile, and the number of matches taken must be a divisor of the number of matches in the other pile. The one who picks up the last match wins. If A takes first and both play optimally, who will win?
［Solution］ B must win. For each natural number $n$, let
$$
v(n)=\max \left\{k\left|k \in \mathbb{N}, 2^{k}\right| n\right\},
$$

and call $v(n)$ the index of $n$. Note that $v(100)=2=v(252)$. We first prove the following lemma.

Lemma If the number of matches in two piles are $n$ and $m$ respectively, and $v(n)=v(m)$, then the first player, after taking $r$ matches from one pile (without loss of generality, assume the first pile), must have $v(n-r) \neq v(m)$.
Proof of the lemma Since $r \mid m$, it follows that
$$
v(r) \leqslant v(m)=v(n).
$$

If $v(r) < v(n) = v(m)$, then $v(n-r) = v(n) = v(m)$, which contradicts the assumption. If $v(r) = v(n) = v(m)$, then $v(n-r) < v(n) = v(m)$, which also contradicts the assumption. Therefore, $v(n-r) \neq v(m)$.

Returning to the solution of the original problem. Since $v(100)=v(252)=2$, by the lemma, after A takes the matches, the indices of the two piles of matches must be unequal. Then B can take matches from the pile with the higher index, and make the indices of the two piles equal after taking. This is always possible. In fact, if the number of matches in the two piles after A takes is $n_{1}$ and $m_{1}$ respectively, and $v\left(n_{1}\right)>v\left(m_{1}\right)$, then B can take $2^{v\left(m_{1}\right)}$ matches from the first pile of $n_{1}$ matches. It is easy to see that after taking, the indices of the two piles are equal. Clearly, as long as B adheres to this strategy, B will eventually win.