# Trailing Zeroes

Let's say you have this question:

## How many trailing zeroes does 100! have?

### Count the tens

The first method that comes to mind is to count the tens or count the 10's and the 2*5's.

1 **2 **3 4 **5 **6 7 8 9 **10**, so in 10! there are 2 zeroes. The problem with that method is it takes a lot of time, and it's completely unnecessary, because we only did the first 10, and you have 90 more numbers to look at.

### Count the fives

An important idea to understand is that **5 is a limiting factor**. In other words, there are fewer 5's than 2's. That's important because in order to multiply to 10 you need a 2 and a 5.

The reason there are more 2's is that every other number has 2 as a factor. Every fifth number has a 5 as a factor. So, we can say that for every 5 in the factorial, there's a 2. So, all we have to do is count the 5's, and we can count the 0's.

So, what you have to do is count the numbers with one 5 (e.g. 5,10,15,20,30,35,40, 45, 55, etc. ), then add 1 for each number with two 5s (e.g. 25, 50, 75, 100), then add 1 for each number with three 5s (e.g. 125, 250, 375, 500), etc. So, our answer is:

### 100/5 + 100/25 + 100/125 = 20 + 4 + 0 (there's no 125 in the first 100 positive integers) = 24

So, there are **24 **trailing zeroes in 100!

## How many trailing zeroes does 133! have?

This is similar, but I just want to demonstrate that you only take the integer part of each quotient.

133/5 + 133/25 + 133/125 = 26.6 + 5.32 + 1.064 (ignore the decimals) = 26 + 5 + 1 = **32**