International Mathematical Olympiad

If $p=2$ or $p=5$ then Eduardo chooses $i=0$ and $a_0=0$ in the first move, and wins, since, independently of the next moves, $M$ will be a multiple of $10$.

Now assume that the prime number $p$ does not belong to $\{2,5\}$. Eduardo chooses $i=p-1$ and $a_{p-1}=0$ in the first move. By Fermat's Little Theorem, ${\left(10^{(p-1)/2}\right)}^2=10^{p-1}\equiv 1(\mathrm{mod}p)$, so $p\mid {\left(10^{(p-1)/2}\right)}^2-1=\left(10^{(p-1)/2}+1\right)\left(10^{(p-1)/2}-1\right)$. Since $p$ is prime, either $p\mid 10^{(p-1)/2}+1$ or $p\mid 10^{(p-1)/2}-1$. Thus we have two cases:

Case a: $10^{(p-1)/2}\equiv -1(\mathrm{mod}p)$

In this case, for each move $\left(i,a_i\right)$ of Fernando, Eduardo immediately makes the move $\left(j,a_j\right)=\left(i+\frac{p-1}{2},a_i\right)$, if $0⩽i⩽\frac{p-3}{2}$, or $\left(j,a_j\right)=\left(i-\frac{p-1}{2},a_i\right)$, if $\frac{p-1}{2}⩽i⩽p-2$. We will have $10^j\equiv -10^i(\mathrm{mod}p)$, and so $a_j\cdot 10^j=a_i\cdot 10^j\equiv -a_i\cdot 10^i(\mathrm{mod}p)$. Notice that this move by Eduardo is always possible. Indeed, immediately before a move by Fernando, for any set of the type $\{r,r+(p-1)/2\}$ with $0⩽r⩽(p-3)/2$, either no element of this set was chosen as an index by the players in the previous moves or else both elements of this set were chosen as indices by the players in the previous moves. Therefore, after each of his moves, Eduardo always makes the sum of the numbers $a_k\cdot 10^k$ corresponding to the already chosen pairs $\left(k,a_k\right)$ divisible by $p$, and thus wins the game.

Case b: $10^{(p-1)/2}\equiv 1(\mathrm{mod}p)$

In this case, for each move $\left(i,a_i\right)$ of Fernando, Eduardo immediately makes the move $\left(j,a_j\right)=\left(i+\frac{p-1}{2},9-a_i\right)$, if $0⩽i⩽\frac{p-3}{2}$, or $\left(j,a_j\right)=\left(i-\frac{p-1}{2},9-a_i\right)$, if $\frac{p-1}{2}⩽i⩽p-2$. The same argument as above shows that Eduardo can always make such move. We will have $10^j\equiv 10^i(\mathrm{mod}p)$, and so $a_j\cdot 10^j+a_i\cdot 10^i\equiv \left(a_i+a_j\right)\cdot 10^i=9\cdot 10^i(\mathrm{mod}p)$. Therefore, at the end of the game, the sum of all terms $a_k\cdot 10^k$ will be congruent to $$\sum _{i=0}^{\frac{p-3}{2}}9\cdot 10^i=10^{(p-1)/2}-1\equiv 0(\mathrm{mod}p)$$ and Eduardo wins the game.