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.
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You may notice that $$P=\frac{200!}{2^{100}\cdot 100!} $$ Now use that a prime $p$ occurs in $n!$ with multiplicity exactly $$\lfloor n/p\rfloor + \lfloor n/p^2\rfloor + \lfloor n/p^3\rfloor + \lfloor n/p^4\rfloor +\ldots $$
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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: $$ \frac{200-6}{3-1} - \frac{100-4}{3-1} = 97-48 = 49 $$
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.