Question

The number of positive integers $n$ such that $2^n$ divides $n!$ is

• A. one
• B. two
• C. infinite
• D. zero

Answer 1

Answer: (B)

Let $E=\frac{2^n}{n!}$
When $n=1$
$E=\frac{2}{1!}=2$
When $n=2$
$E=\frac{2^2}{2!}=\frac{4}{2}=2$
When $n=3$
$E=\frac{2^3}{3!}=\frac{4}{3}$
Hence, two values of $n$ exist.