8.21 Two players, A and B, take turns removing stones from a pile of $n$ stones. A starts first, and on his first turn, he can take any number of stones but not all and not zero. Thereafter, each player must take a number of stones that is a divisor of the number of stones the other player took on their last turn. The player who takes the last stone wins. For what smallest $n>1992$ does B have a winning strategy?
[Solution] First, we assert that for $n>1$, player B has a winning strategy if and only if $n=2^{m}$. Therefore, the smallest $n$ in the problem is 2048.

Let $n=2^{m}, m \in \mathbb{N}$, and suppose player A takes $2^{m} m_{1}\left(2 k_{1}+1\right)$ stones on the first turn, where $m_{1}, k_{1}$ are non-negative integers. Then player B can win by following this strategy: he takes $2^{m_{1}}$ stones on his first turn, and in subsequent turns, he takes the same number of stones as player A just took. In this way, when player A takes stones on the $h$-th turn $(h \geqslant 2)$, the number of stones taken must be a power of 2, say $2^{m_{h}}$. Thus, we must have $0 \leqslant m_{h} \leqslant m_{h-1}$, and after player A's turn, the number of stones left in the pile must be an odd multiple of $2^{m_{h}}$, and certainly not 0. Therefore, the final winner must be player B.

For $n=2^{m}(2 k+1), k \in \mathbb{N}$, player A can ensure victory by taking $2^{m}$ stones on the first turn, leaving $2^{m+1} k$ stones, which B cannot take all at once. Each time B takes stones, he can only take a power of 2 with an exponent not exceeding $m$, leaving an odd multiple of a power of 2. Then player A can adopt the strategy used by B in the previous scenario to ensure a win.