Find the largest $k$ such that $3^k$ divides the product of the first $100$ odd integers 
Let $P$ be the product of the first 100 positive odd integers. Find the largest integer $k$ such that $P$ is divisible by $3^k$.

There are $50$ odd numbers and $50$ even numbers between $0$ and $100$, $99$ being the $50$th odd. So it makes sense that, $199$ is the $100$th odd number.
I got that
$$P = 1 \cdot 3 \cdots 199 = \frac{200!}{2 \cdot 4 \cdot 6 \cdots 198},$$
but that doesn't really help. 
I am after
$$\prod_{n = 0}^{99} (2n + 1),$$
but I don't see a general formula. 
 A: You may notice that $$P=\frac{200!}{2^{100}\cdot 100!} $$
Now use that a prime $p$ occurs in $n!$ with multiplicity exactly $$\lfloor n/p\rfloor + \lfloor n/p^2\rfloor  + \lfloor n/p^3\rfloor + \lfloor n/p^4\rfloor +\ldots  $$
A: How many integers less than $199$ does $3$ divide? Every second of those are odd, starting with $3,9,15,...$
How many integers less than $199$ does $3^2$ divide? Every second of those are odd, starting with $9,27,45,...$ Note that all of those were already divisible by $3$ so already counted once before.
Combine those informations (add up all those figures) up to $3^4$, since $3^5=243>199$. I get $k=33+11+4+1=49$.

This figure is confirmed by Wolfram|Alpha.
A: evidently the highest power of $3$ dividing the first $100$ odd numbers is the same as the highest power of $3$ which divides $\frac{200!}{100!}$ 
$$
200 = 2.3^4 +1.3^3 + 1.3^2+ 0.3^1+2.3^0
$$
so the sum-of-digits in the 3-ary represention of 200 is 6. similarly for 100 the sum of 3-ary digits is 4.
hence the number required is:
$$
\frac{200-6}{3-1} - \frac{100-4}{3-1} = 97-48 = 49
$$
