Prove that 33! Is divisible by `2^(19)` and what is the largest integer n such that 33! Is divisible by `2^(n)`?

Explanation

To solve the problem of proving that \( 33! \) is divisible by \( 2^{19} \) and finding the largest integer \( n \) such that \( 33! \) is divisible by \( 2^n \), we will use the formula for finding the highest power of a prime \( p \) that divides \( n! \):

\[ \text{Power of } p \text{ in } n! = \sum_{k=1}^{\infty} \left\lfloor \frac{n}{p^k} \right\rfloor \]

### Step-by-Step Solution:

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