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HMMT 十一月 2008 · 冲刺赛 · 第 3 题

HMMT November 2008 — Guts Round — Problem 3

专题
Contest Math / 竞赛数学
难度
L3
来源
HMMT

题目详情

  1. [ 5 ] What is the largest x such that x divides 24 · 35 · 46 · 57?
    1 HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUNDst
    10710 7

英文原题

[ 5 ] What is the largest x such that x 2 divides 24 · 35 · 46 · 57?
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1 st HARVARD-MIT NOVEMBER TOURNAMENT, 8 NOVEMBER 2008 — GUTS ROUND

解析

英文解析

  1. [ 5 ] What is the largest x such that x divides 24 · 35 · 46 · 57?
    4 2 2
    Answer: 12 We factor the product as 2 · 3 · 5 · 7 · 19 · 23. If x divides this product, x can have atmost 2 factors of 2, 1 factor of 3, and no factors of any other prime. So 2 · 3 = 12 is the largest value 2
    of x .
    1 HARVARD-MIT NOVEMBER TOURNAMENT, 8 SATURDAY 2008 — GUTS ROUNDst
    10710 7