Belarusian Mathematical Olympiad

Answer: Bob wins.

It is well-known that a natural number $n$ is divisible by $11$ if and only if $(S_o-S_e)÷11$, where $S_o,S_e$ are the sums of the digits on the odd and even positions respectively in the decimal representation of $n$.

Let Ann use the numbers $a$, $c$, and $y$ for her first, second, and third moves respectively; let Bob use the numbers $b$ and $x$ for his first and second moves respectively.

The following cases are possible after two Ann's moves and one Bob's move:

1. $a$ and $c$ occupy two odd positions, $b$ occupies the even position of the asterisks; 2. $a$ and $c$ occupy two even positions, $b$ occupies the odd position of the asterisks (different from the first position); 3. $a$ and $c$ occupy one even and one odd positions, $b$ occupies the even position of the asterisks.

Let $X=\{x_1,x_2,\dots ,x_7\}$ be the set of unused digits before the second Bob's move.

In the first case Bob replaces the asterisk on the even position by $x$. Then $S_o=a+c+y$, $S_e=b+x$. Let $r=b-a-c$. Now the congruence $S_o\equiv S_e(\mathrm{mod}11)$ is equivalent to the congruence $y\equiv x+r(\mathrm{mod}11)$.

If $r\equiv 0(\mathrm{mod}11)$, then $B$ can use any digit from the set $X$ as the number $x$ because the congruence $y\equiv x(\mathrm{mod}11)$ is impossible if $x$ and $y$ are the distinct digits.

Let $r\not\equiv 0(\mathrm{mod}11)$. Suppose that for any digit $x_i\in X$ there exists a digit $y_i\ne x_i$, $y_i\in X$ such that $y_i\equiv x_i+r(\mathrm{mod}11)$, $i=1,2,\dots ,7$. Summing these seven congruences we obtain $$y_1+y_2+\cdots +y_7\equiv x_1+x_2+\cdots +x_7+7r(\mathrm{mod}11).$$ Since it is evident that $y_1+y_2+\cdots +y_7=x_1+x_2+\cdots +x_7$, we see that the obtained congruence is equivalent to the congruence $7r\equiv 0(\mathrm{mod}11)$, which is impossible since $r\not\equiv 0(\mathrm{mod}11)$. Therefore, there exists a number $x\in X$ such that $y\not\equiv x+r(\mathrm{mod}11)$ for any digit $y\in X$ different from $x$. So Bob can use this number $x$ for his second move to win.

Consider two other cases. In these cases Bob replaces the asterisk on the odd position (different from the first one) by $x$. Then we have $$S_o-S_e=(b+x+y)-(a+c)(1)$$ for the second case, and $$S_o-S_e=(c+x+y)-(a+b)(2)$$ or $$S_o-S_e=(a+x+y)-(c+b)(3)$$ for the third one.

Set $r=a+c-b$ in (1), $r=a+b-c$ in (2), and $r=b+c-a$ in (3).

Therefore, the congruence $S_o\equiv S_e(\mathrm{mod}11)$ is equivalent to the congruence $x+y\equiv r(\mathrm{mod}11)$ for all cases. Suppose that for any digit $x_i\in X$ there exists a digit $y_i\in X$ such that $x_i+y_i\equiv r(\mathrm{mod}11)$, $y_i\ne x_i$, $i=1,2,\dots ,7$. It follows that all digits from $X$ are partitioned into the pairs of distinct digits with the sum which is congruent to $r$ modulo $11$, but it is impossible since the number of digits in $X$ is odd. Therefore, there exists a number $x\in X$ such that $y+x\not\equiv r(\mathrm{mod}11)$ for any digit $y\in X$ different from $x$. So Bob can use this number $x$ for his second move to win.