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HMMT 十一月 2018 · GEN 赛 · 第 5 题

HMMT November 2018 — 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 smallest positive integer n for which √ √ √ √ 100 + n + 100 − n is an integer.
解析
  1. Compute the smallest positive integer n for which √ √ √ √ 100 + n + 100 − n is an integer. Proposed by: Michael Tang Answer: 6156 √ √ √ √ The number 100 + n + 100 − n is a positive integer if and only if its square is a perfect square. We have (√ ) √ √ 2 √ √ √ √ √ √ 100 + n + 100 − n = (100 + n ) + (100 − n ) + 2 (100 + n )(100 − n ) √ = 200 + 2 10000 − n. To minimize n, we should maximize the value of this expression, given that it is a perfect square. For √ √ this expression to be a perfect square, 10000 − n must be an integer. Then 200 + 2 10000 − n is √ 2 even, and it is less than 200 + 2 10000 = 400 = 20 . Therefore, the greatest possible perfect square √ 2 value of 200 + 2 10000 − n is 18 = 324 . Solving √ 200 + 2 10000 − n = 324 for n gives the answer, n = 6156 .