2004 AIME II Problems/Problem 8
Problem
How many positive integer divisors of are divisible by exactly 2004 positive integers?
Solution
The prime factorization of 2004 is . Thus the prime factorization of is .
We can count the number of divisors of a number by multiplying together one more than each of the exponents of the prime factors in its prime factorization. For example, the number of divisors of is .
A positive integer divisor of will be of the form . Thus we need to find how many satisfy
We can think of this as partitioning the exponents to and . So let's partition the 2's first. There are two 2's so this is equivalent to partitioning two items in three containers. We can do this in ways. We can partition the 3 in three ways and likewise we can partition the 167 in three ways. So we have as our answer.
See also
2004 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 7 |
Followed by Problem 9 | |
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