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HMMT 二月 2016 · 团队赛 · 第 1 题

HMMT February 2016 — Team Round — Problem 1

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 25 ] Let a and b be integers (not necessarily positive). Prove that a + 5 b 6 = 2016.
    nc
解析

英文解析

  1. [ 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) forthis to be possible. Thus a , b are divisible by 3. But then we get 3 | 2016, which is a contradiction.3
    One can also solve the problem in the same manner by taking modulo 7 exist, since all cubes are 0 or
    ± 1 modulo 7. The proof can be copied literally, noting that 7 | 2016 but 7 - 2016.3
    nc