How to find out $ k $ in this problem If $ k $ is an non negative integer, such that $ 24^k $ divides $ 13! $, what is $ k $ then?
Explanation needed, as I am new to factorials.

Hint:
$$24^k=(2^33)^k=2^{3k}3^k$$
\begin{align}
13! &= (13)(12)(11)(10)(9)(8)(7)(6)(5)(4)(3)(2)\\
&=2^2(3)(2)(3^2)(2^3)(2)(3)(2^2)(3)(2)M \\
&= 2^{10}3^5M
\end{align}
where $M$ is not divisible by $2$ or $3$.
Edit:
For $24^k$ to divide $13!$, we need $3k \leq 10$ (inspect power of $2$) and $k \leq 5$ (inspect power of $3$).
Hence $k \leq \frac{10}{3}$ and $k \leq 5$.
Hence $0 \leq k \leq 3$.
