Indian Mathematicians Aryabhatta and Brahmagupta are given credit for the discovery of Zero. Aryabhatta has given the decimal system and Brahmagupta is given credit for the usage of zero and its operations.
The place of Zero as a digit matters a lot. If it is placed before any number it does not have any impact but at the same point if it is placed after the number it increases the valuation.
To understand the concept of trailing zeros in a better way the following example will help-
02 and 20
In this article, I will discuss the count of ending zeroes. It is impossible to count the exact number of zeroes without actually doing the calculations. So, we are going to understand the concept and the method to Count the number of trailing zeroes or ending zeroes from the prime factorisation of a number and from the factorial of a number.
What is Trailing Zero / Ending zero?
Zeroes are present at the end of the non-zero digits of any number.
NOTE: The prime numbers which are essential for the formation of Zeros are 2 and 5 only.
Finding trailing zeroes from prime factorisation of a number.
Let N be any Natural Number and can be represented in the prime factorisation form
( N = 25 x 33 x 53 ) where the number of trailing zeroes of the number be 3.
we can observe that there are three ( 3 ) pairs of 2 and 5, N = 22 x 33 x (23 x 53)
Thus, N = 108000.
Problem 01:
Find the number of zeroes at the end of 623 x 759 x 1052
Solution:
To get the number of zeroes we have to find out the power raised to 2 and 5 first then notice the pair formed by them.
Let the number N = 623 x 759 x 1052.
N = (223x 323) x ( 3 x 25)9 x ( 5 x 3 x 7)2
N = (223x 323) x ( 3 x 52 )9 x ( 5 x 3 x 7)2
N = 223x 323 +9 + 2 x 52 + 18 = 223x 334 x 520
We observe there are a total number of 20 pairs formed. Thus the total number of trailing zeroes at the end of the expression is 20.
Trailing Zeroes of the factorial Number
What is factorial N or ( N! )?
It is the product of all positive natural numbers up to N.
N! = N x ( N-1 ) x ( N – 2) x ……x 1
I.e 5! = 5 x 4 x 3 x 2 x 1 = 23 x 31 x 51 = 120, here there is only one pair possible. Hence one trailing zero.
Note: The number of 2’s in the prime factorisation of a number is always higher than the number of 5’s. So, if we count the number of 5’s, we are done with the counting of trailing zeroes of any factorial numbers.
The following formula will help you to find the number of trailing zeros of n!
No. of Trailing zeroes = [n/5] + [n/25} + [n/125] + …
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Problem 2:
Find the number of trailing zeroes in 100!
Solution:
Using the formula we get the required number.
No. of Trailing zeroes = [100/5] + [100/25] = 20 + 4 = 24.
Alternate approach:
Step 1: Divide the number successively by 5 till the quotient becomes less than 5.
Step 2: Add all the quotients except the number.
Problem 03:
What will be the largest value of n such that 10n divides the product 25 x 33 x 48 x 53 x 67 x 76 x 812 x 99 x 106 x 1512 x 2014 x 2211 x 2515
Solution:
For the number divisible by 10 it must have the same power for 2 and 5.
Power of 2 = 25 + 16+7+36+6 + 28+ 11 = 2109
Power of 5 = 53 +6+12+14+30 = 565
Hence, the maximum power of 10 is 65.
The very next problem will help you to understand the application.
Problem 4:
What is the maximum value of m, if the number N = 90 x 42 x 324 x 55 is divisible by 3m ?
Solution:
As N is divisible by 3, the maximum value of m depends on the highest value of 3 in the product of the number.
N = 90 x 42 x 324 x 55 = 9 x 10 x 6 x 7 x 81 x 4 x 55
Power of 3 = 32 + 1 +4 = 37
Hence, the maximum value of m is 7.
Read More:
NUMBER SYSTEM: Introduction and concept of unit digit
