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**A ****T****railing zero is a zero digit in the representation of a number which has no non-zero digits that are less significant than the zero digit. Put more simply, it is a zero digit with no non-zero digits to the right of it. ****Representation of Trailing Zeros****(i) Number of trailing zeroes in a Product or Expression**

If we look at a number N, such that

N = 170130000 = 17013*10^{4}

Number of trailing zeroes is the **Power of 10** in the expression or in other words, the number of times N is divisible by 10.

For a number to be divisible by 10, it should be divisible by 2 & 5.

Consider a number N, such that**N = 2 ^{a} * 5^{b} * x**

For the number to have a zero at the end, both a & b should be at least 1.

If you want to figure out the exact number of zeroes, you would have to check how many times the number N is divisible by 10.

When I am dividing N by 10, it will be limited by the powers of 2 or 5, whichever is lesser.

Number of trailing zeroes is going to be the power of 2 or 5, whichever is lesser.

Let us look at an example to further illustrate this idea.

**Q.1.** **What is number of trailing zeroes in 12000?****Ans.** 12000 = 2^{5} * 3 * 5^{3}

When I divide it by 10, it would be divisible exactly thrice because I have only three 5s.

In this case, number of 5s has become the limiting factor and so, the power of 5, which is 3 is the answer.**Tip:** The power of 5 will be the limiting factor in most cases of continuous distribution. It will happen because 5 is less likely to occur than 2.

**Q.2.** **Find out the number of zeroes at the end of N?****Ans.** N = 1^{1} * 2^{2} * 3^{3} .... .100^{100}

Looking at the expression, we can say that the power of 5 will be the limiting factor.

All we need to do is to figure out the number of 5s in the expression.

1^{1}, 2^{2}, 3^{3}, 17^{17}, 89^{89},… will not give us any 5s.

5^{5} will give us five 5s.

10^{10} will give us ten 5s.

15^{15} will give us fifteen 5s.

And so on.

So, the total number of 5s that I have is

⇨ 5 + 10 + 15 ....100 = 5(1+2+3...20) =

But I have made a mistake in the above calculation.

I have assumed that 25^{25} will give me twenty-five 5s but that is incorrect.

It is incorrect because 25^{25} = 5^{50} and will actually give me 50 5s.__Other errors are__:

50^{50} will actually give me 100 5s, whereas I have considered only 50 5s.

75^{75} will actually give me 150 5s, whereas I have considered only 75 5s.

100^{100} will actually give me 200 5s, whereas I have considered only 100 5s.

Considering the above, I have made an error of = 25 + 50 + 75 + 100 = 250 5s.

So the total number of 5s that I have are 1050 + 250 = 1300.

So the number of trailing zeroes at the end of the expression is 1300**(ii) Number of trailing zeroes in a factorial (n!)**

Number of trailing zeroes in n! = Number of times n! is divisible by 10 = Highest power of 10 which divides n! = Highest power of 5 in n!

The question can be put in any of the above ways but it can be answered using the simple formula given below:

*{[x] is the greatest integer function. [4.99] = 4, [4.01] = 4, [- 4.99] = 5, [-4.01] = 5}*

The above formula gives us the exact number of 5s in n! because it will take care of all multiples of 5 which are less than n. Not only that it will take care of all multiples of 25, 125, etc. (higher powers of 5).

Tip:Instead of dividing by 25, 125, etc. (higher powers of 5); it would be much faster if you divided by 5 recursively.

**Q.3.** **What is the number of trailing zeroes in 23!****Ans.** [23/5] = 4. It is less than 5, so we stop here.**The answer is 4.** **Q.4.** **What is the number of trailing zeroes in 123!****Ans.** [123/5] = 24

Now we can either divided 123 by 25 or the result in the above step i.e. 24 by 5.

[24/5] = 4. It is less than 5, so we stop here.**The answer is = 24 + 4 = 28****Q.5.** **What is the number of trailing zeroes in 1123!?****Ans.** [1123/5] = 224

[224/5] = 44

[44/5] = 8

[8/5]=1. It is less than 5, so we stop here.**The answer is = 224 + 44 + 8 + 1 =277****FINDING OUT 'N' WHEN NUMBER OF TRAILING ZEROS IS KNOWN****Q.6.** **Number of trailing zeroes in n! is 13. n = ?**

I need to get 13 trailing zeroes which I will definitely get from 65!

But it will have some extra zeroes in the end because of higher powers of 5.

So, I will consider the previous multiple of 5, which in this case is 60.

Trailing zeroes in 60! = [60/5] + [60/25] = 12 + 2 = 14

I got 14 but I want to get 13, so I will consider the previous multiple of 5, which in this case is 55.

Trailing zeroes in 55! = [55/5] + [55/25] = 11 + 2 = 13

So, the valid values of n! are

Trailing zeroes in 95! = [95/5] + [95/25] = 19 + 3 = 22 {

As you can see from above, we would end up in a loop.

This will happen because

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