Factorial  Atreya Roy

Author: Atreya Roy is pursuing his BTech From Kalyani Government Engineering College, Bengal.
In mathematics, the factorial of a nonnegative integer n, denoted by n!, is the product of all positive integers less than or equal to n. For example, 6 ! = 1*2*3*4*5*6 = 720
Keep in Mind :
1! = 1
2! = 2
3! = 6
4!= 24
5!=120
6!= 720
7! = 5040
8! = 40320
Note : The value of 0! = 1
Example : What is the Unit digit of the series : 1! + 2! + 3!+ 4!.... 100!
Solution :
If we look at the series properly, It is a summation of factorials till 100.
So from 5!, all the unit digit of the individual factorials will be 0.
So basically the sum now turns till 4!
1! + 2! + 3!+ 4! = 1+2+6+24 = 33
Hence the Unit digit of the whole series will be 3
Highest Power of a prime in a number
Let us take an example for the following.
What is the highest power of 3 such that the number will divide 12!
If we write 12! , we get 12! = 1*2*3*4*5*6*7*8*9*10*11*12.
So from 3,6,9 and 12 we will get 3s.
Total 3s = 1 from 3,
1 from 6
2 from 9
1 from 12.
So a total of Five 3s in 12!.
But what about bigger numbers ? It will be very difficult to calculate this hence we will follow a simple formula given below.
For any Factorial “F” denoted as F!, the number of primes “p” is found out using :
[F/p] + [F/p^2] + [F/p^3]…… and so on
Where [ ] denotes the greatest integer function.
Example : What is the highest power of 5 such that the number divided 125! ?
Solution :
[125/5] + [125/25] + [125/125] = 25+5+1 = 31
Highest power of 5 in 125! = 31
Hence 5^31 divided 125!
Example :
What is the highest power of 12 contained in 30!
Solution:
12= 2*2*3
So we need to find the number of 2s and 3s in 30!
2s :
[30/2] + [30/4] + [30/8] + [30/16]
= 15+7+3+1= 26
There are 26 2s , means there are 13 4s
Now 3s:
[30/3] + [30/9] + [30/27]
= 10+ 3+1 = 14
There are 14 3s
So highest power of 12 cannot be more than 13 since we have no more 4s to club with 3s to get 12s. Hence highest power of 12 in 30! = 13.
Number of Zeroes in a factorial
To find the number of zeroes in a factorial, find the number of 2s and 5s in the number
Example :
What is the Number of zeroes at the end of 30!
Solution :
2s :
[30/2] + [30/4] + [30/8] + [30/16]
= 15+7+3+1= 26
5s :
[30/5] + [30/25]
= 6+1 = 7
There are only 7 5s but 26 2s , but to get a 0, we need one 2 and one 5. So maximum zeroes possible = 7 (because we only have 7 5s)
Hence Highest power of 10 in 30! = 7