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

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.

See Also

2008 iTest (Problems)
Preceded by:
Problem 28
Followed by:
Problem 30
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