HMMT 二月 2008 · COMB 赛 · 第 2 题

HMMT February 2008 — COMB Round — Problem 2

[ 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). Determine the average of m ( A ), when A is taken over all nonempty subsets of S .

解析

[ 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). Determine the 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 , 2 2009 which is the constant . 2 Remark: : This argument is very analogous to the famous argument that Gauss used to sum the series 1 + 2 + · · · + 100.