8. Find the largest positive integer $x$ such that 2017 ! is divisible by $19^{x}$.
Answer: 111
Solution: There are $\left\lfloor\frac{2017}{19}\right\rfloor=106$ numbers from 1 to 2017 which are divisible by 2017 . Among these, five numbers, $19^{2}, 2 \cdot 19^{2}, 3 \cdot 19^{2}, 4 \cdot 19^{2}, 5 \cdot 19^{2}$, are divisible by $19^{2}$. Therefore, $19^{111} \mid 2017$ ! and so the largest possible $x$ is 111 .