$n$ divides a product of $n$ consecutive integers (by $n$ divides one of them)
Note that there is some $k$ such that $0\leq k< x$ and $a\equiv k\mod x$. Then $$a+(x-k)\equiv k+x-k\equiv x\equiv 0\mod x$$ and what do you get when you multiply this by other stuff?
One of the numbers $a+1, a+2,\ldots, a+x$ is congruent to $0$ mod $x$. Multiplying by $0$ yields $0$.