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HMMT 二月 2003 · 代数 · 第 6 题

HMMT February 2003 — Algebra — Problem 6

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. Let a = 1, and let a = b n /a c for n > 1. Determine the value of a .
    1 n n − 1 999
    3 2 3 3 3
解析

英文解析

  1. Let a = 1, and let a = b n /a c for n > 1. Determine the value of a .
    1 n n − 1 999
    Solution: 999
    We claim that for any odd n , a = n . The proof is by induction. To get the basen
    3 3
    cases n = 1 , 3, we compute a = 1, a = b 2 / 1 c = 8, a = b 3 / 8 c = 3. And if
    1 2 3
    3 2
    the claim holds for odd n ≥ 3, then a = b ( n + 1) /n c = n + 3 n + 3, so a =
    n +1 n +2
    n +3 n +22
    3 2 3 2 2
    b ( n +2) / ( n +3 n +3) c = b ( n +6 n +12 n +8) / ( n +3 n +2) c = b n +2+ c = n +2.
    n +3 n +32
    So the claim holds, and in particular, a = 999.
    3 2 3 3 3999