<p><strong><span style="color:#e74c3c;"><span style="font-size:18px;">But what if we're not dividing by a prime number?</span></span></strong></p>
<p>For example, what if the question asks us to find the powers of $15$ in $200!$?</p>
<p>$$\frac{200!}{15^x}$$</p>
<p>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:</p>
<p>$$\frac{200!}{(3 \times 5)^x} = \frac{200!}{3^x5^x}$$</p>
<p>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...</p>
<p style="text-align: center;">$5 = $ <span style="color:#e74c3c;">limiting factor</span></p>
<p>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 $200!$.</p>
<p>$$\frac{200!}{5^x}$$</p>
<p style="text-align: center;"><span style="color:#e74c3c;"><em>Continuously divide by $5$ and take the whole number result </em></span></p>
<p>$$40...8...1...0$$</p>
<p><span style="color:#27ae60;">$$40+8+1=49$$</span></p>