2017 AIME II Problems/Problem 8

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$\textbf{Problem 8}$ Find the number of positive integers $n$ less than $2017$ such that \[1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}\] is an integer.

$\textbf{Problem 8 Solution}$ $\boxed{134}$