HMMT 二月 2014 · 冲刺赛 · 第 4 题

英文解析

4. [ 4 ] Let D be the set of divisors of 100. Let Z be the set of integers between 1 and 100, inclusive. Markchooses an element d of D and an element z of Z uniformly at random. What is the probability thatd divides z ?
   2 2217
   Answer: As 100 = 2 · 5 , there are 3 · 3 = 9 divisors of 100, so there are 900 possible pairsof d and z that can be chosen.900
   If d is chosen, then there are possible values of z such that d divides z , so the total number of valid 100
   ∑ ∑d
   100 2 2
   pairs of d and z is = d = (1 + 2 + 2 )(1 + 5 + 5 ) = 7 · 31 = 217. The answer isd | 100 d | 100
   217 dtherefore .
   900