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HMMT 十一月 2013 · 冲刺赛 · 第 20 题

HMMT November 2013 — Guts Round — Problem 20

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [ 11 ] There exist unique nonnegative integers A, B between 0 and 9, inclusive, such that 2 (1001 · A + 110 · B ) = 57 , 108 , 249 . Find 10 · A + B .
解析
  1. [ 11 ] There exist unique nonnegative integers A, B between 0 and 9, inclusive, such that 2 (1001 · A + 110 · B ) = 57 , 108 , 249 . Find 10 · A + B . 2 2 Answer: 75 We only need to bound for AB 00; in other words, AB ≤ 5710 but ( AB + 1) ≥ 5710. A quick check gives AB = 75. (Lots of ways to get this...)