2004 AIME II Problems/Problem 8
Problem
How many positive integer divisors of 
are divisible by exactly 2004 positive integers?
Solution 1
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.
Solution 2 (bash)
Clearly we need to find a group of numbers that multiply to 2004. We can list them all out since we know that 2004 is only 
.
167, 2, 2, 3
4, 3, 167
12, 167
4, 501
2, 1004
2, 3, 334
2, 2, 501*
6, 2, 167
3, 668
6, 334
2004*
To begin, the first multiple doesn't work because there are only 3 prime divisors of 2004. We can apply all multiples because the prime factorization of 
is 
. Every multiple has six ways of distributing numbers to become powers of 167, 3, and 2, except for the ones with a star.
For a single power of 2004, we have three choices (2, 3, and 167) to give a power of 2003 to.
For 2, 2, 501, there are three choices to give a power of 500 to and the rest get a power of 1.
Therefor we have 
normal multiples and 
"half" multiples. Sum to get 
as our answer.
See also
| 2004 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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