# How many zeros are there at the end of 100 factorial?

How many zeros at the end of 100 factorial? How many trailing zeros in 1000 factorial? Do you get baffled by questions like these?

### How useful is calculator in solving this problem?

Sadly, calculator is not of much help because the answer will be in scientific notation. I will get the first few digits but what I am really bothered about in this case is the end of the multiplied expansion which the calculator will not be able to display.

This trailing zeros in a factorial is fairly simple to handle once you are equipped with the right way to handle it and then you will be able to handle 3468796! with equal ease as 10!.

There are many interesting derivation from factorials.

## How to quickly find number of zeros at the end of factorial of any number or say ‘n!’?

Let us get a few facts straight.

First, any number will have a zero at the end if that number is a multiple of 10; or in other words the number has 10 as a factor. So basically I have to find out how many 10s are there in the expansion of n!

Since 5X2=10, so I will have to consider all the products of 5 and 2 because each such product will give me a 10.

Obviously there are way more multiples of 2 than multiples of 5 in any given factorial expansion. This is because every 5th number is a multiple of 5 whereas every 2nd number is a multiple of 2.

Therefore, to find the number of 10s in the expansion of n!, what I am really bothered about is how many times 5 is a factor in all the numbers between 1 and n.

If the above logic is clear, we can go through an example for further clarity.

### Figuring out the number of 5s in a factorial

Find the number of trailing zeros in 17!

17! = 17x16x15x14x13x12x11x10x9x8x7x6x5x4x3x2x1

Basically I have to take all the numbers with 5 as a factor which when clubbed with a multiple of 2 will yield a 10. multiples of 5 are 15, 10 and 5.

Hence, 17! has three trailing zeros.

Let us now look at a few more examples.

## How many zeros at the end of 100 factorial?

Multiples of 5 in the expansion of 100! are 5, 10, 15, 20,….., 100.

You can count it this way and waste a lot of time or consider this.

100 divided by 5 equals 20. So there are 20 multiples of 5 from 1 to 100.

But wait, 25=5X5. This means that 25 is giving me an extra 5 which I need to account for. With this logic every multiple of 25 will give me an extra 5.

100 divided by 25 equals 4. So there are 4 multiples of 25 from 1 to 100.

Therefore number of trailing zeros in 100! = 100/5+100/25= 20+4 = 24

The same reasoning extends to larger numbers.

## How many Zero at the end of 1000 factorial?

There are 1000/5=200 multiples of 5 between 1 and 1000.

The next power of 5 is 25 and there are 1000/25=40 multiples of 25 between 1 and 1000.

The next power of 5 is 5^{3}=125 and each such multiple of 125 gives me still one more extra 5 as compared to other multiples of 25. There are 1000/125 = 8 such multiples of 125 between 1 and 1000.

The next power of 5 is 5^{4}=625 and there is only one multiple of 625 between 1 and 1000, 625 itself.

Hence the total number of 5s in the expansion of 1000! = 200 + 40 + 8 + 1 = 249 and thus 249 trailing zeros in 1000!

## How to find zeros in any factorial?

Take the number.

Divide by 5; if you get a decimal consider only the whole number part.

Divide by 5^{2}=25; if you get a decimal consider only the whole number part.

Divide by 5^{3}=125; if you get a decimal consider only the whole number part.

Divide by 5^{4}=625; if you get a decimal consider only the whole number part.

Continue this division with higher powers of 5 until your division results in a number less than one that is only a fraction with no whole number part.

Add up all the whole numbers you got in the series of divisions you performed. This is the number of trailing zeros. Since you’ve learned the method of finding trailing zeros for any factorial, it will be easy for you to find the largest power of a number that divides a factorial number

One last example before I wind up.

### Find the number of trailing zeros in 3426!

While performing the divisions, I am only considering the whole number part and conveniently ignoring the decimal or fraction part.

5^{1}: 3426/5= 685

5^{2}: 3426/25= 137

5^{3}: 3426/125= 27

5^{4}: 3426/625= 5

5^{5}: 3426/3125= 1

So the total number of trailing zeros is 685+137+27+5+1 = 855.

By the way you can get the same result by dividing the number repeatedly by 5 till you get 1 as quotient and then adding up all the results. This saves a lot of time as you don’t have to deal with division by higher powers of 5.

3426 ÷ 5 = 685

685 ÷ 5 = 137

137 ÷ 5 = 27

27÷ 5 = 5

5 ÷ 5 = 1

685 + 137 + 27 + 5 + 1 = 855 which is the number of trailing zeros in 3426!

Hope this article was of help to you. See you with many more such ways to tackle seemingly difficult problems. You can post your queries in the comment box below or post it on Quickermaths.com/questions. Till then, happy solving! 🙂

Excellent and very useful for the students appearing for Bank exams and other competitive exams.

Thanks for the appreciation R. Rama Rao. Will try to post more such helpful posts! 🙂