But what if we're not dividing by a prime number?
For example, what if the question asks us to find the powers of $15$ in $200!$?
$$\frac{200!}{15^x}$$
We cannot simply use the trick we learned previously because $15$ is not a prime number. Thus, we need to rewrite the problem like so:
$$\frac{200!}{(3 \times 5)^x} = \frac{200!}{3^x5^x}$$
So now we're dealing with prime numbers. But the question is do we focus on the $3$ or the $5$? Well, to make one $15$, we need one $3$ and one $5$. Which is more common? In $200!$, are we going to find more powers of $3$ or more powers of $5$? We're going to find more powers of $3$. That means that...
$5 = $ limiting factor
So really this problem is asking us to focus solely on the powers of $5$ given that they are fewer in number than the powers of $3$. To find the number of powers of $15$ in $200!$, we simply find the number of powers of $5$ in $200!$.
$$\frac{200!}{5^x}$$
Continuously divide by $5$ and take the whole number result
$$40...8...1...0$$
$$40+8+1=49$$