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?
$\textbf{(A) } 0 \qquad \textbf{(B) } 1\qquad\textbf{(C) } 2\qquad\textbf{(D) } 3\qquad\textbf{(E) } 4$
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 $\boxed{\textbf{(E) } 4}$ ways.
~MRENTHUSIASM