Problem:
Let a1,a2,…,a2018 be a strictly increasing sequence of positive integers such that
a1+a2+⋯+a2018=20182018
What is the remainder when a13+a23+⋯+a20183 is divided by 6?
Answer Choices:
A. 0
B. 1
C. 2
D. 3
E. 4
Solution:
Let n be an integer. Because n3−n=(n−1)n(n+1), it follows that n3−n has at least one prime factor of 2 and one prime factor of 3 and therefore is divisible by 6. Thus n3≡n(mod6). Then
a13+a23+⋯+a20183≡a1+a2+⋯+a2018≡20182018(mod6)
Because 2018≡2(mod6), the powers of 2018 modulo 6 are alternately 2,4,2,4,…, so 20182018≡4(mod6). Therefore the remainder when a13+a23+⋯+a20183 is divided by 6 is (E)4.
The problems on this page are the property of the MAA's American Mathematics Competitions