AMC8 2022 8, Problem 3

When three positive integers $a$, $b$, and $c$ are multiplied together, their product is $100$. Suppose $a < b < c$. In how many ways can the numbers be chosen?

Solution: The positive divisors of $100$ are \[1,2,4,5,10,20,25,50,100.\] It is clear that $10\leq c\leq50,$ so we apply casework to $c:$ If $c=10,$ then $(a,b,c)=(2,5,10).$ If $c=20,$ then $(a,b,c)=(1,5,20).$ If $c=25,$ then $(a,b,c)=(1,4,25).$ If $c=50,$ then $(a,b,c)=(1,2,50).$ Together, the numbers $a,b,$ and $c$ can be chosen in $\fbox{4}$ ways.

Answer: 4