Prove that 33! is divisible by `" 2^15`. what is the largest integer n such that 33! is divisible by `2^n`.

Explanation

For prime number `p` and `n in N`, the highest power of `p` that divides `n!` is given by,

`sum_(k ge 1) [n/p^k]`.

Here, `n = 33` and `p = 2`.
So, highest power that divides `33! = [33/2]+[33/2^2]+[33/2^3]+[33/2^4]+[33/2^5]`

`=16+8+4+2+1 = 31`

So, `33!` can be divided by `2^31`.

So, `33!` is divisible by `2^15` and maximun value of `n` is `31` such that `33!` is divisible by `2^31`.