HMMT 二月 2013 · 冲刺赛 · 第 7 题

HMMT February 2013 — Guts Round — Problem 7

[ 5 ] Find the number of positive divisors d of 15! = 15 · 14 · · · · · 2 · 1 such that gcd( d, 60) = 5.

解析

[ 5 ] Find the number of positive divisors d of 15! = 15 · 14 · · · · · 2 · 1 such that gcd( d, 60) = 5. i ′ Answer: 36 Since gcd( d, 60) = 5, we know that d = 5 d for some integer i > 0 and some integer ′ ′ d which is relatively prime to 60. Consequently, d is a divisor of (15!) / 5; eliminating common factors ′ 2 with 60 gives that d is a factor of (7 )(11)(13), which has (2 + 1)(1 + 1)(1 + 1) = 12 factors. Finally, i can be 1 , 2 , or 3, so there are a total of 3 · 12 = 36 possibilities.