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HMMT 二月 2006 · GEN1 赛 · 第 4 题

HMMT February 2006 — GEN1 Round — Problem 4

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. Find √ √ √ 31 + 31 + 31 + . . . √ . √ √ 1 + 1 + 1 + . . .
解析
  1. Find √ √ √ 31 + 31 + 31 + . . . √ . √ √ 1 + 1 + 1 + . . . √ Answer: 6 − 5 2 Solution: Let the numerator be x and the denominator y . Then x = 31 + x , so, as x > 0, we have √ √ 1 + 1 + 4 · 31 1 + 5 5 x = = . 2 2 Similarly we compute that √ √ 1 + 1 + 4 · 1 1 + 5 y = = , 2 2 so that √ √ √ √ √ x 1 + 5 5 1 + 5 5 1 − 5 − 24 + 4 5 √ √ √ = = · = = 6 − 5 . y − 4 1 + 5 1 + 5 1 − 5 1