. The first digit is written by the first player, the second by the second player. The third digit is again written by the first player, and so on. Can the first player ensure that the resulting number is divisible by 9, if the second player wants to prevent this? Consider the cases: a) $k=10$; b) $k=15$.
Let the given number $N=\overline{a_{1}^{1}} \overline{a_{2}^{2}} \overline{-} \overline{-} \overline{2 k-1} \overline{-} \bar{a}_{2 k}$, where $a_{i}$ is one of the digits $1,2,3,4,5$, and the digits with odd indices are chosen by the first player (A), while the digits with even indices are chosen by the second player (B). Let $S_{i}=a_{1}+a_{2}+\ldots+a_{i}$. The number $N$ is divisible by 9 if and only if $S_{2 k}$ is divisible by 9.

Suppose $k$ is divisible by 3. We will prove that in this case, player B wins. For any move $a_{2 i-1}$ by player A, he should respond with the move $a_{2 i}=6-a_{2 i-1}$. In this case, $S_{2 i}=6 i$ for any $i \leq k$, in particular, $S_{2 k}=6 k$ is divisible by 9.

Now suppose $k$ is not divisible by 3. We will show that in this case, player A wins. For this, he should choose $a_{1}=3$, and then apply the strategy of B in the first case, that is, for any move $a_{2 i}$ by player B, respond with the move $a_{2 i+1}=6-a_{2 i}$. In this case, $S_{2 k}=3+6(k-1)+a_{2 k}=6 k-3+a_{2 k}$. If $k=3 n+1$, then $S_{2 k}=18 n+3+a_{2 k}$ gives a remainder of $3+a_{2 k} \leq 8$ when divided by 9. If $k=3 n+2$, then the remainder of $S_{2 k}=18 n+9+a_{2 k}$ when divided by 9 is $a_{2 k} \neq 0$. In each of these cases, the number $S_{2 k}$ is not divisible by 9.

## Answer

a) Can; b) cannot.