jmc

$\boxed{\boxed{11}\times \boxed{20}}$ $11$ has $2$ factors, and $20$ has $\{1,2,4,5,10,20\}$ as factors, for a total of $6$ factors. Plugging $\boxed{11}=2$ and $\boxed{20}=6$: $\boxed{2\times 6}$ $\boxed{12}$ $12$ has factors of $\{1,2,3,4,6,12\}$, so $\boxed{12}=6$, and the answer is $\boxed{6}$