# CAT Exam Notes: Concepts and Important Questions of Factorials

** 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.**

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.

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## Important Questions of Factorials for CAT Exam

**Find the highest power of a prime number in a given factorial**

The highest power of prime number p in n! = [n/p^{1}] + [n/p^{2}] + [n/p^{3}] + [n/p^{4}] + …

**Find the highest power of 3 in 100!**

= [100/3] + [100/3^{2}] + [100/3^{3}] + [100/3^{4}] + [100/3^{5}] + …

= 33 + 11 + 3 + 1 + 0 (from here on greatest integer function evaluates to zero)

= 48.

**What is the greatest power of 5 which can divide 80! exactly** (Another way of asking what is the highest power of 5 in 80!)

= [80/5] + [80/5^{2}] + [80/5^{3}] + …

= 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.

**What is the greatest power of 30 in 50!**

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).**

**What is the greatest power of 8 in 50!**

8 =2^{3}

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 p**^{a}** in n! = [Highest power of p in n!]/a**

**n! in its standard form**

**Find the highest power of all prime numbers, less than n, in n!**

**What are the number of factors of 15!**

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! = 2^{11 }* 3^{6 }* 5^{3 }* 7^{2}* 11^{1 }* 13^{1}

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.

**Find the number of zeros at the end of 100!**

**Find the number of zeros at athe end of 100! in base 7**

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.**

So how many trailing zeros are there in 500! in base 9 ?

**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 it is

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!)^{3}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.**

For 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…

We saw Triangular numbers where instead of multiplying the numbers, we added n consecutive numbers… Remember why it is called as Triangular numbers ?

The first triangular number is 1.

The second one is 1 + 2 = 3.

The third triangular number is 1 + 2 + 3 = 6 and so on.

We can calculate the nth triangular number using the formula **[(n) (n+1)]/2**

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