A Trailing zero is a zero digit in the representation of a number which has no nonzero digits that are less significant than the zero digit. Put more simply, it is a zero digit with no nonzero 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 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 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 = ?
Ans. 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!
Q.7. Number of zeroes in n! is 23. n =?
Ans. 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.
Use Code STAYHOME200 and get INR 200 additional OFF

Use Coupon Code 
75 videos87 docs156 tests
