It's alright to count

The highest power of $2$ that divides $n!$ is given by:

$\begin{aligned} p(n) & = \sum_{k=1}^\infty \left \lfloor \frac{n}{2^k} \right \rfloor \end{aligned}$

where $\lfloor x \rfloor$ is the greatest integer function.

For $n=33$ , we have:

$\begin{aligned} p(33) & = \sum_{k=1}^\infty \left \lfloor \frac{33}{2^k} \right \rfloor \\ & = \left \lfloor \frac{33}{2} \right \rfloor + \left \lfloor \frac{33}{4} \right \rfloor + \left \lfloor \frac{33}{8} \right \rfloor + \left \lfloor \frac{33}{16} \right \rfloor + \left \lfloor \frac{33}{32} \right \rfloor \\ & = 16+8+4+2+1 \\ & = \boxed{31} \end{aligned}$

$\nu_2(33!)=\nu_2(32!)=\frac{32-1}{2-1}=\boxed{31}$ by Legendre's Theorem

The highest power of p dividing any number a is given by:

[a/p] + [a/p^2] +...

use it for a = 33 and p = 2

It's not as fancy or as ‘clean’ as how other people solved it, but an intuitive way to approach the problem (for those who don't know/understand math terminology as well)

I just considered each number in 33! and calculated how many times you could divide each by 2. 2/2 = 1, 3/2= 0, 4/2= 2, 5/2 = 0, 6/2= 1, etc… And then added those numbers up to get 31.

33/2 +33/4+33/8+33/16+33/32=16+8+4+2+1=31

Solved it computationally. Basically, compute n! mod 2^k (for k = 0..2n, say).

I used the brute force method. Enter 33! into calc divided it by 2 until I got an odd number. I hit the return key 31 times.