## What is the largest non-negative integer $k$ such that ${24}^{k}$ divides 13! ? Thank you!

Nov 9, 2016

$k = 3$

#### Explanation:

Using the properties of exponents that ${\left(a b\right)}^{x} = {a}^{x} {b}^{x}$ and ${\left({a}^{x}\right)}^{y} = {a}^{x y}$, we have

${24}^{k} = {\left({2}^{3} \cdot {3}^{1}\right)}^{k} = {\left({2}^{3}\right)}^{k} \cdot {\left({3}^{1}\right)}^{k} = {2}^{3 k} \cdot {3}^{k}$

Thus 13! is divisible by ${24}^{k}$ if and only if 13! is divisible by ${2}^{3 k}$ and is divisible by ${3}^{k}$.

We can tell the greatest power of $2$ by which 13! is divisible by if we look at its factors which are divisible by $2$:

$2 = {2}^{1}$
$4 = {2}^{2}$
$6 = {2}^{1} \cdot 3$
$8 = {2}^{3}$
$10 = {2}^{1} \cdot 5$
$12 = {2}^{2} \cdot 3$

As none of the odd factors contribute any factors of $2$, we have

13! = (2^1*2^2*2^1*2^3*2^1*2^2)*m = 2^(10)*m

where $m$ is some integer not divisible by $2$. As such, we know that 13! is divisible by ${2}^{3 k}$ if and only if ${2}^{10}$ is divisible by ${2}^{3 k}$, meaning $3 k \le 10$. As $k$ is an integer, this means $k \le 3$.

Next, we can look at which factors of 13! are divisible by $3$:

$3 = {3}^{1}$
$6 = {3}^{1} \cdot 2$
$9 = {3}^{2}$
$12 = {3}^{1} \cdot 4$

As no other factors of 13! contribute any factors of $3$, this means

13! = (3^1*3^1*3^2*3^1)*n = 3^5*n

where $n$ is some integer not divisible by $3$. As such, we know that ${3}^{5}$ is divisible by ${3}^{k}$, meaning $k \le 5$.

The largest nonnegative integer satisfying the constraints $k \le 3$ and $k \le 5$ is $3$, giving us our answer of $k = 3$.

A calculator will verify that (13!)/24^3 = 450450, whereas (13!)/24^4=18768.75