Number of Trailing Zeros  Ravi Handa

In this article, I would like to cover these two ideas:
 Number of trailing zeroes in an expression
 Number of trailing zeroes in n!
But before I begin, let us first try to understand what exactly are ‘trailing zeroes’. It is nothing else but the number of zeroes at the end. I do not want to sound pedantic but on many occasions when you see a question which asks about, “What is the number of zeroes in ___” it is incorrect, because it should actually say – “What is the number of trailing zeroes?” or “What is the number of ending zeroes?” It is virtually impossible to predict the exact number of zeroes without actually doing the calculation and finding out the answer.
Just to clarify, 170130000 has 5 zeroes but 4 trailing/ending zeroes.In questions based on these ideas, you should assume that the examiner is asking about trailing zeroes unless specified otherwise.
Number of trailing zeroes in an 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.
Eg1: What is number of trailing zeroes in 12000?
12000 = 25 x 3 x 53
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.Eg2: Find out the number of zeroes at the end of 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) = 5 * (20 * 21) / 2 = 1050But I have made a mistake in the above calculation.
I have assumed that 25^25 will give me twentyfive 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 1300Number 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:
[n/5] + [n/5^2] + [n/5^3] + ...{[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.
Let us use this to solve a few examples:
Eg1: What is the number of trailing zeroes in 23! ?
[23/5 ]= 4. It is less than 5, so we stop here.
The answer is 4.Eg2: What is the number of trailing zeroes in 123! ?
[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 = 28Eg3: What is the number of trailing zeroes in 1123!?
[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 =277Eg4: Number of trailing zeroes in n! is 13. n =?
There is no standard formula for such type of questions but they can be solved by a little bit of hit and trial.
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 55!, 56!, 57!, 58!, 59!Eg5: Number of zeroes in n! is 23. n =?
Trailing zeroes in 100! = [100/5] + [100/25 ] = 20 + 4 = 24 {Too high. Consider previous multiple}
Trailing zeroes in 95! = [95/5] + [95/25] = 19 + 3 = 22 {Too low. Consider next multiple}
As you can see from above, we would end up in a loop.
This will happen because there is no valid value of n for which n! will have 23 zeroes in the end.