2008 iTest Problems/Problem 29

Problem

Find the number of ordered triplets $(a,b,c)$ of positive integers such that $abc=2008$ (the product of $a, b$ , and $c$ is $2008$ ).

The number $2008$ can be factored into $2^3 \cdot 251$ . Use casework to organize the counting.

If two numbers are $1$ , then the third one must be $2008$ , and there are $3$ ways to write the ordered pairs.
If one number is a $1$ , then there are $\tfrac{8-2}{2} = 3$ possible pairs of numbers for the other two. Since the numbers are all different, there are $3 \cdot 6 = 18$ ways to write the ordered pairs.
If none of the numbers are $1$ , then since there are only four prime numbers being multiplied, one of the numbers must have two prime numbers being multiplied together. Thus, the two sets of numbers are $2,2,502$ and $2,4,251$ , and there are $3+6=9$ ways in this case.

Altogether, there are $\boxed{30}$ ordered pairs that satisfy the criteria.