【Question 30】
There are two piles of stones, one with 11 stones and the other with 13 stones, denoted as $(11,13)$. Two players, A and B, take turns to remove stones, following these rules: one can take any number of stones from one pile or the same number of stones from both piles, but must take at least one stone. The player who takes the last stone wins. Now, A goes first. When A first takes to $(\ldots)$, A is guaranteed to win.
【Analysis and Solution】Strategic problem.

We denote the state where taking $(m, n)$ leads to a win or a guaranteed winning strategy as $\star$.

Obviously, $(0,0)$ is $\star$;
Then, $(0, a)(a \neq 0)$, all with a difference of 0 are not $\star$.
$(1,2)$ can only turn into $(0,2) 、(1,1) 、(0,1)$; and $(0,2) 、(1,1) 、(0,1)$ are all not $\star$, meaning $(1,2)$ cannot turn into $(0,0)$, hence $(1,2)$ is $\star$; thus, $(1, b)(b \neq 2),(2, c)(c \neq 1)$, all with a difference of 1 are not $\star$.

Similarly, $(3,5)$ cannot turn into $(0,0),(1,2)$, so $(3,5)$ is $\star$;
By analogy, $(0,0),(1,2),(3,5),(4,7),(6,10),(8,13),(9,15),(11,18),(12,20),(14,23), \ldots$ are all $\star$; therefore, when Player A first takes $(3,5)$ or $(8,13)$, Player A is guaranteed to win.