jmc

Since $12^n=2^{2n}\cdot 3^n$, we are looking for the largest value of $n$ such that $2^{2n}$ and $3^n$ are divisors of $20!$. $$\frac{20}{2}=10\frac{10}{2}=5\frac{5}{2}=2.5\frac{2}{2}=1$$ The largest power of 2 that divides $20!$ is $2^{(10+5+2+1)}=2^{18}$. $$\frac{20}{3}=6\frac{2}{3}\frac{6}{3}=2$$ The largest power of 3 that divides $20!$ is $3^{(6+2)}=3^8$. Since there are 18 powers of 2 and 8 powers of 3 in $20!$, we want the largest value of $n$ such that $2n\le 18$ and $n\le 8$, so $\boxed{8}$ is the answer and $12^8$ is the largest power of 12 that divides $20!$.