Problem 2332

Assume that $19$ does not divide the product $a_{1} a_{2} \cdots a_{9}$. This means that $a_{1}, a_{2}, \ldots, a_{9}$ are relatively prime with $19$. Using Fermat,
$$a_{1}^{18} \equiv a_{2}^{18} \equiv \cdots \equiv a_{9}^{18} \equiv 1 \pmod{19}.$$
But $a_{i}^{18} \equiv 1 \pmod{19}$ is equivalent to $(a_{i}^{9} - 1)(a_{i}^{9} + 1) \equiv 0 \pmod{19}$. Since $19$ is a prime number, this implies that $a_{i}^{9} \equiv \pm 1 \pmod{19}$, for $i = 1, \ldots, 9$, and therefore $a_{1}^{9} + a_{2}^{9} + \cdots + a_{9}^{9} \equiv \pm k \pmod{19}$ for some odd number $k$ between $1$ and $9$. This is a contradiction. Thus $19$ divides the product $a_{1} a_{2} \cdots a_{9}$.