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