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HMMT 二月 2003 · 冲刺赛 · 第 7 题

HMMT February 2003 — Guts Round — Problem 7

专题
Discrete Math / 离散数学
难度
L3
来源
HMMT

题目详情

  1. [6] a and b are integers such that a + b = 15 + 216. Compute a/b .
解析
  1. a and b are integers such that a + b = 15 + 216. Compute a/b . Solution: 1 / 2 √ √ 2 Squaring both sides gives a + b +2 a b = 15+ 216; separating rational from irrational 2 2 2 parts, we get a + b = 15 , 4 a b = 216, so a and b equal 6 and 9. a is an integer, so √ 2 a = 9 , b = 6 ⇒ a/b = 3 / 6 = 1 / 2. (We cannot have a = − 3, since a + b is positive.)