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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