# 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 $2008$).

## Solution 1

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.

## Solution 2 $2008$ can be prime factorized into $2^3\cdot251$. We can think of each ordered pair $(a,b,c)$ as a way to assign three 2s and one 251 to three distinct letters. You may now recognized this as a "assign non-distinct objects to distinct piles" problem. In problems like this, we should use stars and bars. There are $\binom{5}{2}=10$ ways to assign three 2s to three distinct letters, and there are $\binom{3}{2}=3$ ways to assign one 251 to three distinct letters. Multiplying, we get $3\cdot10=\boxed{30}$.