<s>[INST] 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.
 [/INST] The highest power of a prime $p$ in $n!$ is given by $$\sum_{k=1} \Biggl \lfloor  \frac{n}{p^k}\Biggr \rfloor  $$ 
$24^k = (2^3 \times 3)^k = 2^{3k} \times 3^k$
Highest Power of $2$ in $13!$
Let the highest power be $x$. 
$x= \lfloor \frac{13}{2} \rfloor  + \lfloor \frac{13}{2^2} \rfloor   + \lfloor \frac{13}{2^3} \rfloor  = 10 $
Similarly 
Highest power of $3$ in $13!$ =$ 5$
Since $24^k= 2^{3k} \times 3^k$ 
$0 \le k \le \frac{10}{3}$  (Since $2^{3k}$ is present in $24^k$ and maximum power of $2$ is $10$)
and $0 \le k \le 5$ (Since maximum power of $3$ in $13!$ is $5$)
Combining the above two inequalities 
$0 \le k \le 3$ 
So $k=0,1,2,3$
You can read more about the formula used here
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