Greece — Selection Examination

$$M = a_0 + a_1 p + \dots + a_{p-1} 10^{p-1},$$
where $a_i$ is the value that the $i$-th box will take (from right to left).
If $p = 2$ or $p = 5$, then Angelo puts $0$ in the last box and wins.
If $p \neq 2, 5$, then by Fermat's Theorem we have
$$p|10^{p-1}-1 \Leftrightarrow p|(10^{\frac{p-1}{2}})^2-1 \Leftrightarrow p|(10^{\frac{p-1}{2}}-1)(10^{\frac{p-1}{2}}+1)$$
so $p|(10^{\frac{p-1}{2}}-1)$ or $p|(10^{\frac{p-1}{2}}+1)$. We have two cases:

(α) If $p|(10^{\frac{p-1}{2}}+1)$. Then at the first move Angelo plays zero at $a_{p-1}$. Then, each time that Vangelis puts $a_i$ at the $i$-th position, Angelo plays symmetric (i.e. at $j=i+\frac{p-1}{2}$ if $0 \le i \le \frac{p-3}{2}$, or at $j=i-\frac{p-1}{2}$ if $\frac{p-1}{2} \le i \le p-2$) putting the digit $a_i$ at the $j$-th position.
Then,
$$10^j \equiv -10^i \pmod{p} \Rightarrow a_j 10^j = a_i 10^j \equiv -a_i 10^i \pmod{p}$$
so $p|a_j 10^j + a_i 10^i$. We observe that after Angelo's first move, it remains to fill $p-1$ positions, which is an even number. Matching as above, Angelo can make the number divisible by $p$.

$$10^j = 10^i (\text{mod } p) \Rightarrow a_j 10^j + a_i 10^i = (a_i + a_j) 10^i = 9 \cdot 10^i (\text{mod } p)$$
In the above matching we get sums of the form $9 \cdot 10^i (\text{mod } p)$, then the total sum equals
$$M = \sum_{i=0}^{\frac{p-3}{2}} 9 \cdot 10^i = 10^{(p-1)/2} - 1 \equiv 0 (\text{mod } p),$$
and thus Angelo wins.

(β) If $p|(10^{\frac{p-1}{2}}-1)$. Then, at the first move Angelo plays $0$ at $a_{p-1}$. Then, each time that Vangelis puts $a_i$ at the $i$-th position, Angelo plays symmetric with respect to the center of the number (i.e. $j=i+\frac{p-1}{2}$ if $0 \le i \le \frac{p-3}{2}$, or at $j=i-\frac{p-1}{2}$ if $\frac{p-1}{2} \le i \le p-2$) putting the digit $9-a_i$ at $j$-th position.