It is trivially done using Legendre method $([120/2]+[120/4]+[120/8]..+1)/2 =58$ is the largest power of 4 dividing 120! Similarly, $([120/3]+[120/9]...=58$ is the largest power of 3 in 120!. Hence 12^58 divides $120!$ Note: [X] is Greatest integer function

58 find the power of 3 in 120

Let's define the arithmetic function $e_{l}(t)$ as the highest power of the number $l$ dividing $t$. Then by Legendre's theorem we,ve $e_{p}(k!)=\sum_{k \ge 1} \left \lfloor \frac{k}{p^{k}} \right \rfloor $ where $p$ is a prime.Then $e_{3}(120!)=\sum_{k \ge 1} \left \lfloor \frac{120}{3^{k}} \right \rfloor =58$. and $e_{2}(120!)=\sum_{k \ge 1} \left \lfloor \frac{120}{2^{k}} \right \rfloor =116$ thus $e_{4}(120!)=116/2=58$. As $e_{lm}(t)=\max\{e_{l}(t),e_{m}(t)\}$, when $gcd(l,m)=1$ we can conclude the $e_{12}(120!)=58$

$k_{max}=58$ Highest power of any positive integer $n$ in $r!$ is $n^{[\frac{r}{3}] + [\frac{r}{3^2}]+... [\frac{r}{3^x}]}$ here $[y]$ is the greatest integer $<y$

This problem seemed too easy for a P3. By Legendre's the number of factors of $2$ in $120!$ is $116$ and the number of factors of $3$ in $120!$ is $58$. Since $12=2^2 \cdot 3$ so we can have at most $\min(116/2, 58/1)=\boxed{58}$