Question

The largest integer $n$ such that $33!$ is divisible by $2^n$ is

• A. $30$
• B. $31$
• C. $32$
• D. None of thes

Answer 1

The correct option is B $31$

In terms of prime factors $33!$ can be written as $2^a.3^d.5^c.7^d....$

$E_2(33!)=\frac{33}{2}+\frac{33}{2^2}+\frac{33}{2^3}+\frac{33}{2^4}+\frac{33}{2^5}+\frac{33}{2^6}=16+8+4+2+1=31$

Hence the eexponential of $2$ in $33!$ is $31$ so largest value of $n$ is 31.