Question

The largest non-negative integer $k$ such that ${24}^k$ divides $13!$ is

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

The correct option is B $3$

${24}^k=2^{3k}\times 3^k$
Exponent of $2$ in $13!$
$=\left[\frac{13}{2}\right]+\left[\frac{13}{2^2}\right]+\left[\frac{13}{2^3}\right]+\left[\frac{13}{2^4}\right]+\cdots$
$=6+3+1=10$
(Here, [ ] denotes greatest integer function)
Exponent of $3$ in $13!$
$=\left[\frac{13}{3}\right]+\left[\frac{13}{3^2}\right]+\left[\frac{13}{3^3}\right]+\cdots$
$=4+1=5$
$\Rightarrow$If ${24}^k$ divides $13!,$ then the highest possible exponent of $2\text{and}3$ can be $10\text{and}5$ respectively.
$\Rightarrow 3k\le 10\text{and}k\le 5$
$\therefore k=3$