HMMT 二月 2016 · 团队赛 · 第 1 题

HMMT February 2016 — Team Round — Problem 1

[ 25 ] Let a and b be integers (not necessarily positive). Prove that a + 5 b 6 = 2016. c n

解析

[ 25 ] Let a and b be integers (not necessarily positive). Prove that a + 5 b 6 = 2016. Proposed by: Evan Chen 3 3 Since cubes are 0 or ± 1 modulo 9, by inspection we see that we must have a ≡ b ≡ 0 (mod 3) for 3 this to be possible. Thus a , b are divisible by 3. But then we get 3 | 2016, which is a contradiction. One can also solve the problem in the same manner by taking modulo 7 exist, since all cubes are 0 or 3 ± 1 modulo 7. The proof can be copied literally, noting that 7 | 2016 but 7 - 2016. c n