HMMT 十一月 2012 · 团队赛 · 第 2 题

英文解析

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.
   3 2
   Answer: 800 Since 360 = 2 · 3 · 5, the only possible prime divisors of d are 2, 3, and 5, so weia b ci i ican write d = 2 · 3 · 5 , for nonnegative integers a , b , and c . Then, d d d | 360 if and only if thei 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 assignmentsiof b that satisfy the second inequality, and 4 assignments of c that satisfy the third inequality, for ai itotal 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
   integer quadruplets ( a , a , a , a ) satisfying a + a + a + a = n .)3
   1 2 3 4 1 2 3 4