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HMMT 十一月 2020 · GEN 赛 · 第 5 题

HMMT November 2020 — GEN Round — Problem 5

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Compute the sum of all positive integers a ≤ 26 for which there exist integers b and c such that a + 23 b + 15 c − 2 and 2 a + 5 b + 14 c − 8 are both multiples of 26.
解析
  1. Compute the sum of all positive integers a ≤ 26 for which there exist integers b and c such that a + 23 b + 15 c − 2 and 2 a + 5 b + 14 c − 8 are both multiples of 26. Proposed by: David Vulakh Answer: 31 Solution: Assume b and c exist. Considering the two values modulo 13, we find { a + 10 b + 2 c ≡ 2 (mod 13) 2 a + 5 b + c ≡ 8 (mod 13) . Subtracting twice the second equation from the first, we get − 3 a ≡ − 14 (mod 13). So, we have a ≡ 9 (mod 13). Therefore we must either have a = 9 or a = 22. Moreover, both a = 9 and a = 22 yield solutions with b = 0 and c = 3 , 16, depending on the value of a . Thus the answer is 9 + 22 = 31.