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HMMT 十一月 2012 · 团队赛 · 第 4 题

HMMT November 2012 — Team Round — Problem 4

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 π be a permutation of the numbers from 2 through 2012. Find the largest possible value oflog π (2) · log π (3) · · · log π (2012).
    2 3 2012
解析

英文解析

  1. [ 3 ] Let π be a permutation of the numbers from 2 through 2012. Find the largest possible value oflog π (2) · log π (3) · · · log π (2012).
    2 3 2012
    Team Round
    Answer: 1 Note that
    2012 2012
    ∏ ∏
    log π ( i )
    log π ( i ) =
    log iii =2 i =2
    ∏
    2012
    log π ( i )
    i =2
    ∏=
    2012
    log ii =2 = 1 ,
    where the last equality holds since π is a permutation of the numbers 2 through 2012.