Preselection tests for the full-time training

Assume that $19$ does not divide the product $a_1{a}_2\cdots a_9$. This means that $a_1,a_2,\dots ,a_9$ are relatively prime with $19$. Using Fermat, $$a_1^{18}\equiv a_2^{18}\equiv \cdots \equiv a_9^{18}\equiv 1(\mathrm{mod}19).$$ But $a_i^{18}\equiv 1(\mathrm{mod}19)$ is equivalent to $(a_i^9-1)(a_i^9+1)\equiv 0(\mathrm{mod}19)$. Since $19$ is a prime number, this implies that $a_i^9\equiv \pm 1(\mathrm{mod}19)$, for $i=1,\dots ,9$, and therefore $a_1^9+a_2^9+\cdots +a_9^9\equiv \pm k(\mathrm{mod}19)$ for some odd number $k$ between $1$ and $9$. This is a contradiction. Thus $19$ divides the product $a_1{a}_2\cdots a_9$.