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# Factorials - Number System Quant Notes | EduRev

## Albanian : Factorials - Number System Quant Notes | EduRev

The document Factorials - Number System Quant Notes | EduRev is a part of the Albanian Course Quantitative Aptitude.
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Factorial is an important topic in quantitative aptitude preparation not just because of its importance as an independent concept, but also because of its application in many topics like permutation & Combination, Probability etc… The factorial of a non negative integer n is denoted as n! The notation was introduced by Christian Kramp in 1808.
n! is calculated as the product of all positive integers less than or equal to n.
i.e 6! = 1 * 2 * 3 * 4 * 5 * 6 = 720
n! = 1 when n = 0, and n! = (n - 1)! * n if n > 0

➢ What is the significance of n!
"n! is the number of ways we can arrange n distinct objects into a sequence".
Example: 2! = 2 means numbers 1, 2 can be arranged in 2 sequences (1, 2) and (2, 1).
We can arrange 0 in one way. So 0! = 1, not zero. Now we know why, and no need to say “its like that” if someone asks.

Find the highest power of a prime number in a given factorial.
The highest power of prime number p in n! = [n/p1] + [n/p2] + [n/p3] + [n/p4] + …
[x] denotes the greatest integer less than or equal to x.
[1.2] =1
[4] = 4
[-3.4] = -4 (why? Because -4 is less than -3.2)

Example 1: Find the highest power of 3 in 100!
Solution: [100/3] + [100/32] + [100/33] + [100/34] + [100/35] + …
= 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)
= 48.

Example 2: What is the greatest power of 5 which can divide 80! exactly?
Solution: [80/5] + [80/52] + [80/53] + …
= 16 + 3 + 0 + … = 19.

Find the highest power of a non prime number in a given factorial
We learnt before that any non prime number can be expressed as a product of its prime factors. We will use this concept to solve this type of questions.

Example 1: What is the greatest power of 30 in 50!
Solution: 30 = 2 * 3 * 5.
Find what the highest power of each prime is in the given factorial
We have 25 + 12 + 6 + 3 + 1 = 47    2s in 50!
16 + 5 + 1 = 22    3s in 50!
10 + 2 = 12    5s in 50!
We need a combination of one 2, one 3 and one 5 to get 30. In the given factorial we have only 12 5s. Hence only twelve 30s are possible. So the greatest power of 30 in 50! is 12.

Note: We don’t have to find the power of all prime factors of a given number. It is enough to find the greatest power of the highest prime factor in the factorial (here 5).

Example 2: What is the greatest power of 8 in 50!
Solution: 8 =23
50! has 47 twos. We need 3 two to get an 8.
We can have a maximum of [47/3] = 15 8s in 50!

Note: The highest power of the number pa in n! = [Highest power of p in n!]/a.

➢ Find the highest power of all prime numbers, less than n, in n!
Example 1: What are the number of factors of 15!
Solution: 2, 3, 5, 7, 11 and 13 are the prime numbers less than 15.
Highest power of 2 in 15! = 11
Highest power of 3 in 15! = 6
Highest power of 5 in 15! = 3
Highest power of 7 in 15! = 2
Highest power of 11 in 15! = 1
Highest power of 13 in 15! = 1
15! = 211 * 3* 5* 72* 11* 131
Thus we can find the number of divisors, sum of divisors and other values for 15!
number of factors of 15! = (11 +1) (6 + 1) (3 + 1) ( 2 + 1) ( 1 + 1) (1 + 1) = 4032

➢ Find the number of zeros at the end of a factorial value
This is just another way of asking what highest power of 10 is in a given factorial.
10 = 2*5, we need a combination of one 2 and one 5 to get a zero at the end. Just find the number of 5s in the given factorial as number of 2s will be always more than number of 5s.

Example 1: Find the number of zeros at the end of 100!
Solution: [100/5] + [100/25] + … = 20 + 4 + 0 + … = 24.

Example 2: Find the number of zeros at at the end of 100! in base 7
Solution: Concept is same as before. in base 10 we need a 2 and 5 to get a zero but in base 7 we need a 7 to get a zero. so we have to find the highest power of 7 in 100!
[100/7] + [100/49] + ... = 14 + 2 = 16. There will be 16 zeros at the end of 100! in base 7

Note:  To find the number of zeros of a factorial in a base b, calculate the highest power of b in that factorial. b need not be prime, we already know how to find the highest power of a composite number in a factorial expression.

WILSON'S THEOREM
For every prime number p, (p-1)! + 1 is divisible by p
Take p =7, then as per Wilson's theorem 7 is prime only if (6!) + 1 is divisible by 7 and some useful results derived from Wilson's theorem.

Remainder [(p-1)!/p] = p - 1 ( where p is a prime number )
Remainder [6!/7] = 7; Remainder [100!/101] = 100 and so on...

Remainder [(p-2)!/p] = 1 ( where p is a prime number )
Remainder [5!/7] = 1; Remainder [99!/101] = 1 and so on...

SOME EXTRA CONCEPTS
➢ Product of n consecutive natural numbers is divisible by n!
11 * 12 * 13 * 14 is completely divisible by 4!
This can be extended to obtain other interesting results.
Consider 1 * 2 * 3 * 4 * 5 * 6 . as per above result 1 * 2 * 3 * 4 * 5 * 6 is divisible by 6!.
Now take (1 * 2 * 3) * (4 * 5 * 6) . each of this is divisible by 3!
so 1 * 2 * 3 * 4 * 5 * 6 is completely divisible by (3!)2
similarly 1 * 2 * 3 * 4 * 5 * 6 is completely divisible by (2!)also.

Note: To generalise (a*n)! is completely divisible by (n!)a,  where a and n are integers.

➢ The product of all odd (even) integers up to an odd (even) positive integer n is called the double factorial of n. (though it has only about half the factors of the original factorial!). It is denoted as n!!
9!! = 1 * 3 * 5 * 7 * 9 = 945
8!! = 2 * 4 * 6 * 8 = 384

➢ Product of first n! is called the super factorial of n, denoted as sf(n).
sf(4) = 1! * 2! * 3! * 4!

➢ A factorion is a natural number that equals the sum of the factorials of its decimal digits.
Example: 145 is a factorion because 1! + 4! + 5! = 1 + 24 + 120 = 145.

Note: There are just four factorions (in base 10) and they are 1, 2, 145 and 40585.

➢ A factorial prime is a prime number that is one less or one more than a factorial.
2 = 1! + 1, 3 = 2! + 1, 5 = 3! -1, 7 = 3! + 1 etc…

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## Quantitative Aptitude

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