Huge numbers

Firstly, finding the highest power of $2$ which divides $36!$ ,

$=\left \lfloor \frac{36}{2} \right \rfloor+\left \lfloor \frac{36}{2^2} \right \rfloor+\left \lfloor \frac{36}{2^3} \right \rfloor+\left \lfloor \frac{36}{2^4} \right \rfloor+\left \lfloor \frac{36}{2^5} \right \rfloor \\ = 18+9+4+2+1=34$

So, $4^{17}$ also divides $36!$ (this result will be helpful for us later in the solution).

Thus, option having $2^{38},2^{36}$ and $24^{36}$ are rejected.

Now, we can see that highest power $17$ is $2$ which can divide $36!$ .

Now, let's find the highest power of $3$ which can divide $36!$ ,

$=\left \lfloor \frac{36}{3} \right \rfloor+\left \lfloor \frac{36}{3^2} \right \rfloor+\left \lfloor \frac{36}{3^3} \right \rfloor \\ =12+4+1=17$

As $4^{17}$ and $3^{17}$ both divide $36!$ , therefore, $12^{17}$ also divides $36!$ .