03/17/2021 (2011: Countdown Round, Problem 24)

Q: What is the number of zeroes to the right of the right-most non-zero number in the product of the first 150 positive integers.

A: 37

Basically, this question is asking for how many factors of 10 are in 150!. To do this, we must find the number of 5's in 150! (because the number of 2's will always outnumber the numbers of 5 in the product).

We find the number of multiples of 5 to be: 150/5 = 30

We find the number of multiples of 25 to be: 150/25 = 6

We find the number of multiples of 125 to be: 150/125 = 1.2 → 1

We add these together to get 30 + 6 + 1 = 37.

Why does this work?

When we divide by 5, we are finding the multiples of 5, but we aren't finding how many 5's 'fit' into each multiple. So then we have to add one for every multiple of 5² to find the additional 5's we missed. And we do this until we reach a 1 or a number that is smaller than 1. We then add the integers greater than 1 that we found.