Problem:
How many positive integer divisors of 20042004 are divisible by exactly 2004 positive integers?
Solution:
A positive integer N is a divisor of 20042004 if and only if N=2i3j167k with 0≤i≤4008,0≤j≤2004, and 0≤k≤2004. Such a number has exactly 2004 positive integer divisors if and only if (i+1)(j+1)(k+1)=2004. Thus the number of values of N meeting the required conditions is equal to the number of ordered triples of positive integers whose product is 2004 . Each of the unordered triples {1002,2,1},{668,3,1},{501,4,1},{334,6,1},{334,3,2},{167,12,1}, {167,6,2}, and {167,4,3} can be ordered in 6 possible ways, and the triples {2004,1,1} and {501,2,2} can each be ordered in 3 possible ways, so the total is 8⋅6+2⋅3=54​.
OR
Begin as above. Then, to find the number of ordered triples of positive integers whose product is 2004 , represent the triples as (2a1​⋅3b1​⋅167c1​,2a2​⋅3b2​⋅167c2​,2a3​. 3b3​⋅167c3​), where a1​+a2​+a3​=2,b1​+b2​+b3​=1, and c1​+c2​+c3​=1, and the ai​ 's, bi​ 's, and ci​ 's are nonnegative integers. The number of solutions of a1​+a2​+a3​=2 is (24​) because each solution corresponds to an arrangement of two objects and two dividers. Similarly, the number of solutions of both b1​+b2​+b3​=1 and c1​+c2​+c3​=1 is (13​), so the total number of triples is (24​)(13​)(13​)=6⋅3⋅3=54​.
The problems on this page are the property of the MAA's American Mathematics Competitions