HMMT 十一月 2012 · 团队赛 · 第 2 题

HMMT November 2012 — Team Round — Problem 2

[ 5 ] Find the number of ordered triples of divisors ( d , d , d ) of 360 such that d d d is also a divisor 1 2 3 1 2 3 of 360.

解析

[ 5 ] Find the number of ordered triples of divisors ( d , d , d ) of 360 such that d d d is also a divisor 1 2 3 1 2 3 of 360. 3 2 Answer: 800 Since 360 = 2 · 3 · 5, the only possible prime divisors of d are 2, 3, and 5, so we i a b c i i i can write d = 2 · 3 · 5 , for nonnegative integers a , b , and c . Then, d d d | 360 if and only if the i i i i 1 2 3 following three inequalities hold. a + a + a ≤ 3 1 2 3 b + b + b ≤ 2 1 2 3 c + c + c ≤ 1 1 2 3 Now, one can count that there are 20 assignments of a that satisfy the first inequality, 10 assignments i of b that satisfy the second inequality, and 4 assignments of c that satisfy the third inequality, for a i i total of 800 ordered triples ( d , d , d ). 1 2 3 (Alternatively, instead of counting, it is possible to show that the number of nonnegative-integer triples ( ) n +3 ( a , a , a ) satisfying a + a + a ≤ n equals , since this is equal to the number of nonnegative- 1 2 3 1 2 3 3 integer quadruplets ( a , a , a , a ) satisfying a + a + a + a = n .) 1 2 3 4 1 2 3 4