Question

The largest non-negative integer $k$ such that ${24}^k$ divides $12!$ is.

• A. $2$
• B. $3$
• C. $4$
• D. $5$

Answer 1

Answer: (B)

$3$

${24}^k{24}^k\to {(2^3\times 3)}^k$
Exponent of $2$ in $13!$
$\left[\frac{13}{2}\right]+\left[\frac{13}{2^2}\right]+\left[\frac{13}{2^3}\right]=10$
Exponent of $3$ in $13!$
$\left[\frac{13}{3}\right]+\left[\frac{13}{3^2}\right]=5$
So ${(2^3\times 3)}^3$ So $K=3$