HMMT 二月 2014 · 冲刺赛 · 第 4 题

HMMT February 2014 — Guts Round — Problem 4

[ 4 ] Let D be the set of divisors of 100. Let Z be the set of integers between 1 and 100, inclusive. Mark chooses an element d of D and an element z of Z uniformly at random. What is the probability that d divides z ? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT 2014, 22 FEBRUARY 2014 — GUTS ROUND Organization Team Team ID#

解析

[ 4 ] Let D be the set of divisors of 100. Let Z be the set of integers between 1 and 100, inclusive. Mark chooses an element d of D and an element z of Z uniformly at random. What is the probability that d divides z ? 217 2 2 Answer: As 100 = 2 · 5 , there are 3 · 3 = 9 divisors of 100, so there are 900 possible pairs 900 of d and z that can be chosen. 100 If d is chosen, then there are possible values of z such that d divides z , so the total number of valid d ∑ ∑ 100 2 2 pairs of d and z is = d = (1 + 2 + 2 )(1 + 5 + 5 ) = 7 · 31 = 217. The answer is d | 100 d | 100 d 217 therefore . 900