This question already has an answer here:
How many trailing zeroes are there in $52!$ ?
I believe I correctly solved it, but I'm not happy with the scalability of my method.
I figure the number of trailing zeroes is equal to the number of times $52!$ is divisible by 10. I wrote out every integer from 1-52 that is divisible by 2 or 5. The idea being that the number of 2 AND 5-factors equals the number of 10-factors.
I quickly noted that the number of 2-factors is greater than the number of 5-factors, so I figured finding the number of 5-factors will do. There were 12.
I'm not very happy with this, because if they now ask me to do the same for $152!$, I'll have to tell them to shove it. I'm not doing this again.
Is there a better way to do this? Perhaps a method that scales better?