Selection Examination

$$M=a_0+a_{1}p+\cdots +a_{p-1}10^{p-1},$$ 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 $p\ne 2,5$, then by Fermat's Theorem we have $$p|10^{p-1}-1\iff p|(10^{\frac{p-1}{2}}{)}^2-1\iff p|(10^{\frac{p-1}{2}}-1)(10^{\frac{p-1}{2}}+1)$$ 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, $$10^j\equiv -10^i(\mathrm{mod}p)\Rightarrow a_{j}10^j=a_{i}10^j\equiv -a_{i}10^i(\mathrm{mod}p)$$ 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.