Sum-thing to Keep You Busy

Since we don't have a reasonable approach to find suitable values of $x$ immediately, the first step is to just pretend like the floor function is nonexistent; doing so gives us $\sum_{n=1}^8 \left\lfloor \frac{x}{n!} \right\rfloor \approx \sum_{n=1}^8 \frac{x}{n!} = x\sum_{n=1}^8 \frac{1}{n!} \approx x(e-1) \approx x1.72.$

Since we're looking for the instances where the sum is about equal to $1729$ or $2015$ , we can roughly estimate that $1729 \le 1.72x \le 2015 \Longrightarrow 1005.2 \le x \le 1171.5.$

Checking close values gives us the actual bounds as $1008 \le x \le 1175$ , so the number of positive integers $x$ that satisfy our equation is $A = 1175-1008+1=168$ .

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Since $168$ factors as $2^3 \cdot 3 \cdot 7$ , the number of factors, $Q$ , is $(3+1)(1+1)(1+1) = 16$ .

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All that remains is evaluating $16^{1729} \bmod 519$ . Using Euler's theorem and the fact that $\phi(519) = 344$ , this expression reduces to $16^{9}$ $\equiv 16((16^2)^2)^2$ $\equiv \fbox{325} \bmod 519$ , our answer.