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HMMT 二月 2008 · COMB 赛 · 第 2 题

HMMT February 2008 — COMB Round — Problem 2

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

英文原题

  1. [ 3 ] Let S = { 1 , 2 , . . . , 2008 } . For any nonempty subset A ⊂ S , define m ( A ) to be the median of A
    (when A has an even number of elements, m ( A ) is the average of the middle two elements). Determinethe average of m ( A ), when A is taken over all nonempty subsets of S .
解析

英文解析

  1. [ 3 ] Let S = { 1 , 2 , . . . , 2008 } . For any nonempty subset A ⊂ S , define m ( A ) to be the median of A
    (when A has an even number of elements, m ( A ) is the average of the middle two elements). Determinethe average of m ( A ), when A is taken over all nonempty subsets of S .
    2009 ′
    Answer: For any subset A , we can define the “reflected subset” A = { i | 2009 − i ∈ A } . Then
    ′ ′2
    m ( A ) = 2009 − m ( A ). Note that as A is taken over all nonempty subsets of S , A goes through all
    ′
    m ( A )+ m ( A )
    the nonempty subsets of S as well. Thus, the average of m ( A ) is equal to the average of ,
    20092
    which is the constant .
    Remark: : This argument is very analogous to the famous argument that Gauss used to sum the series 2
    1 + 2 + · · · + 100.