Question

Prove that $33!$ is divisible by $2^{15}$

Answer 1

Largest power of $2$ in $m!$

$=\left[\frac{m}{2}\right]+\left[\frac{m}{2^2}\right]+\left[\frac{m}{2^3}\right]+.....=\left[\frac{33}{2}\right]+\left[\frac{33}{2^2}\right]+\left[\frac{33}{2^3}\right]+\left[\frac{33}{2^4}\right]+\left[\frac{33}{2^5}\right]=16+8+4+2+1=31$

so $n=31$

$2^{31}$ is divisible by $33!$

so $2^{15}$ is also divisible by $33!$