Problem

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.

Solution

The denominator contains $2,3,5$ . Therefore, one possibility is that $n|30$ . This yields the numbers $30,60,90,120,\cdots,2010$ . There are a total of ${67}$ numbers in the sequence. We express the last two terms as $\frac{6n^5+n^6}{720}\implies\frac{n^5(6+n)}{720}$ This yields that $n \equiv 24,30$ . Therefore, we get the final answer of $\boxed{134}$

2017 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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2017 AIME II Problems/Problem 8

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