HMMT 十一月 2013 · GEN 赛 · 第 6 题

HMMT November 2013 — GEN Round — Problem 6

[ 5 ] Find the number of positive integer divisors of 12! that leave a remainder of 1 when divided by 3. 2 2 2 2

解析

[ 5 ] Find the number of positive integer divisors of 12! that leave a remainder of 1 when divided by 3. 10 5 2 1 1 Answer: 66 First we factor 12! = 2 3 5 7 11 , and note that 2 , 5 , 11 ≡ − 1 (mod 3) while 7 ≡ 1 a b c d (mod 3). The desired divisors are precisely 2 5 7 11 with 0 ≤ a ≤ 10, 0 ≤ b ≤ 2, 0 ≤ c ≤ 1, 0 ≤ d ≤ 1, and a + b + d even. But then for any choice of a, b , exactly one d ∈ { 0 , 1 } makes a + b + d even, so we have exactly one 1 (mod 3)-divisor for every triple ( a, b, c ) satisfying the inequality constraints. This gives a total of (10 + 1)(2 + 1)(1 + 1) = 66. 2 2 2 2