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

${24}^K\to (2^3\times 3{)}^K$

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 $(2^3\times 3{)}^3$

Hence, K=3