2008 iTest Problems/Problem 30

Problem

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

The prime factorization of 2008 is $2^3 \cdot 251$ . Once again, use casework to organize the possibilities.

If $a = 1$ , the number of possibilities of $b$ and $c$ are really the number of distinct pairs of two numbers that multiply to $2008$ that do not have $1$ . In this case, there are $\tfrac{4 \cdot 2}{2} - 1 = 3$ possibilities.
If $a = 2$ , then the only possibility that works is when $b = 4$ and $c = 251$ , for another $1$ possibility.
If $a = 4$ (or higher), then there are no possibilities that work because one of the values of $b$ and $c$ would be smaller than $a$ .

In total, there are $\boxed{4}$ ordered triplets that satisfy the conditions.