How to find out $ k $ in this problem

Question

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.

Answer 1

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 $0 \leq k \leq 3$.

Answer 2

You need $2^{3k}$ and $3^k$ dividing $13!$. Now $$13!=1\times2\times3\times4\times5\times6\times7\times8\times9\times10\times11\times12\times13$$ and the highest power of $3$ dividing this is $$1\times1\times3\times1\times1\times3\times1\times1\times9\times1\times1\times3\times1=3^5.$$ Therefore $k\le 5$. But you also have to take into account powers of $2$. Over to you!

Answer 3

I don't know how to get direct answer but $13!=(4!=\color{green}{24})\times5\times\color{red}{6}\times7\times(8=2\times \color{red}{4})\times 9\times(10=5\times \color{blue}{2})\times 11\times \color{blue}{12}\times 13$

Then $k=0,1,2,3$

Note: I used $(*=*)$ inside that product to reduce number of steps. This is not recommended in any exam etc.

Answer 4

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