GMAT Math Tutorial
Number of zeros at the end of n!
n! is defined as product of 1st n natural numbers.
n! = n*(n-1)*(n-2)*….3*2*1, n must be natural (whole) number.
Eg. 4!= 4*3*2*1 = 24
5!= 5*4*3*2*1 = 120 and so on.
Note : 0! = 1 (explanation lies in theory of permutation & combination)
Solution: To solve this question we need to understand that one pair of 2 & 5 when multiplied gives one zero so in 25! we need to find as how many 2 and 5 are there. And how many pairs of 2 & 5 can be made from that. Off course number of pairs would be decided by that number which is lesser in numbers and in our case number of 5 would be lesser than number of 2. So our task now reduces to find the numbers of 5s only in 25!. To find number of 5s, divide 25 by 5 and get the quotient. Quotient in this case is 5. Divide this quotient again and get the new quotient again and this time quotient is 1. We will keep dividing each quotient till we get the last quotient less than divisor. At this point add all the quotients we get right from the beginning and the sum would be the answer.
Quotient (25/5) = 5
Quotient (5/5) = 1 ( quotient is 1 which is less than 5 so we will stop division at this stage )
So answer is 5+1=6.
Answer for 26!,27!,28!,29! would be same. Think how ?
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