Question

If $33!$ is divisible by $2^n$, then find the maximum value of n.

Answer 1

For prime number $p$ and $n$ the highest power of $p$ that can divide $p!$ can be given as

$\sum \lfloor \frac{n}{p^k}\rfloor n=33,p=2=\lfloor \frac{33}{2^1}\rfloor +\lfloor \frac{33}{2^2}\rfloor +\lfloor \frac{33}{2^3}\rfloor +\lfloor \frac{33}{2^3}\rfloor +\lfloor \frac{33}{2^4}\rfloor +\lfloor \frac{33}{2^5}\rfloor =16+8+4+2+1=31$

So $n=31$