Problem:
Find the number of positive integers n less than 2017 such that
1+n+2!n2+3!n3+4!n4+5!n5+6!n6
is an integer.
Solution:
The sum 2!n2+3!n3+4!n4+5!n5+6!n6 is an integer if and only if multiplying the sum by 6! gives an integer M=360n2+120n3+30n4+6n5+n6 that is divisible by 720.
This means that n6 must be divisible by 6. Suppose n=6k for some positive integer k. Then M=360⋅62k2+120⋅63k3+30⋅64k4+66k5+66k6, which is divisible by 720 if and only if k5+k6 is divisible by 5. This happens exactly when either k or k+1 is a multiple of 5. Thus n must be either a multiple of 30 or 6 less than a multiple of 30. The possible values of n are 30m, where 1≤m≤67, and 30m−6, where 1≤m≤67. There are 134 such values.
The problems on this page are the property of the MAA's American Mathematics Competitions