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.
We start with the last two terms of the polynomial $1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+\frac{n^4}{4!}+\frac{n^5}{5!}+\frac{n^6}{6!}$, which are $\frac{n^5}{5!}+\frac{n^6}{6!}$. This can simplify to $\frac{6n^5+n^6}{720}$, which can further simplify to $\frac{n^5(6+n)}{720}$. Notice that the prime factorization of $720$ is $5\cdot3\cdot3\cdot2\cdot2\cdot2\cdot2$. In order for $\frac{n^5(6+n)}{720}$ to be an integer, one of the parts must divide $5, 3$, and $2$. Thus, one of the parts must be a multiple of $5, 3$, and $2$, and the LCM  of these three numbers is $30$. This means \[n^5 \equiv 0\pmod{30}\] or \[6+n \equiv 0\pmod{30}\] Thus, we can see that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$. Note that as long as we satisfy $\frac{6n^5+n^6}{720}$, $2!, 3!$, and $4!$ will all divide into integers, as their prime factorizations will be fulfilled with the LCM being 30. E.g. $4! = 2\cdot2\cdot2\cdot2\cdot3$, and this will be divisible by $2^4\cdot3^4\cdot5^4$. Now, since we know that $n$ must equal $0\pmod{30}$ or $-6\pmod{30}$ in order for the polynomial to be an integer, $n\equiv0, 24\pmod{30}$. To find how many integers fulfill the equation and are $<2017$, we take $\left \lfloor\frac{2017}{30} \right \rfloor$ and multiply it by $2$. Thus, we get $67\cdot2=\boxed{134}$.
~Solution by IronicNinja~