Question 12
Let $p \geq 2$ be a prime number. Louis and Arnaud take turns choosing an index $i \in \{0,1, \ldots, p-1\}$ that has not yet been chosen, together with a digit $a_{i} \in \{0,1, \ldots, 9\}$. Louis starts. When all indices have been chosen, they compute the following sum:
$$
a_{0}+a_{1} \cdot 10+\ldots+a_{p-1} \cdot 10^{p-1}=\sum_{i=0}^{p-1} a_{i} \cdot 10^{i}
$$
If this sum is divisible by $p$, Louis wins; otherwise Arnaud wins. Show that Louis has a winning strategy.
$$
a_{0}+a_{1} \cdot 10+\ldots+a_{p-1} \cdot 10^{p-1}=\sum_{i=0}^{p-1} a_{i} \cdot 10^{i}
$$
If this sum is divisible by $p$, Louis wins; otherwise Arnaud wins. Show that Louis has a winning strategy.