# Sum-thing to Keep You Busy

Let $$A$$ be the number of positive integers $$x$$ that satisfy

$\displaystyle 1729 \le \sum_{n=1}^8 \left\lfloor \frac{x}{n!} \right\rfloor \le 2015,$

where $$\lfloor x \rfloor$$ denotes the greatest integer less than or equal to $$x$$.

If $$Q$$ is the number of positive integral divisors of $$A$$, find the remainder when $$Q^{1729}$$ is divided by $$519$$.

(You don't have to use a calculator, but it might help)

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