2017 AIME II Problems/Problem 8

Revision as of 12:31, 23 March 2017 by Mewtwomew (talk | contribs) (Solution)

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, $n|30$. This yields the numbers, $30,60,90,120,\cdots,2010$. There are a total of $\boxed{67}$ numbers in the sequence.


EDIT: 0 mod 30 and 24 mod 30 both work; therefore, the answer should be $\boxed{134}$.

See Also

2017 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 7
Followed by
Problem 9
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

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