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FACTORIALS AND THEIR APPLICATIONS

We will see some of the important properties of factorials. A thoroughunderstanding of Factorials is important because they play a pivotal role notonly in the Concepts surrounding Numbers but also other important topics likePermutation and Combination

Definition of Factorial N! = 1*2*3*…*(n-1)*n

Eg) 5! = 1*2*3*4*5=120 3!=1*2*3=6

The applications of Factorials have been categorise d into the following

I) Highest power of a number in a factorial or in a product

II) Number of zeros in the end of a factorial or a product

III) Number of factors of any factorial

I) Highest power of a number in a factorial or in a pr oduct

Questions based on highest power in a factorial are seen year after year in CAT.

What is highest power?

Suppose you have a number N= x2y. Here the highest power of “x” in N will be 2 and the highest power of “y” in N will be 1. Questions based onthis can be categorized based on the nature of the number whose highest power we are finding in the factorial, i.e.

a) Highest power of a prime number in a factorial:

To find the highest power of a prime number (x) in a factorial (N!), continuously divide N by x and add all the quotients.

ILLUSTRATION:Eg 18) The highest power of 5 in 100!

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; ;

Adding the quotients, its 20+4=24. So highest power of 5 in 100! = 24

ALTERNATIVE METHOD

; (We take upto 52 as it is the highest power of 5 which is less than 100)

b) Highest number of a composite number in factorial

1) Factorize the number into primes.

2) Find the highest power of all the prime numbers in that factorial using the previous method.

3) Take the least power.

ILLUSTRATION:

Eg 19) To find the highest power of 10 in 100!

Factorize 10=5*2.

1. Highest power of 5 in 100! =24

2. Highest power of 2 in 100! =97

Therefore, the answer will be 24, because to get a 10, you need a pair of 2 and 5, and only 24 such pairs are available. So takethe lesser value i.e. 24 is the answer.

Eg 20) The highest power of 12 in 100!

12=22 *3

Find the highest power of 22 and 3 in 100!

First find out the highest power of 2.

Listing out the quotients:

Highest power of 2 = 50 + 25 + 12 + 6 + 3 + 1 = 97

So highest power of (out of 97 2’s only 48 can make 22)

Now for the highest power of 3

; Highest power of 3 = 48

Highest power of 12 = 48

II) Number of zeros in the end of a factorial or a prod uct

In base 10, number of zeros in the end depends on the number of 10s; i.e. effectively, on the number of 5s (10=5×2. if you see the previousexample, you will get a lesser number of 5s than 2s. that is why we need to calculate only the number of 5s)

In base N, number of zeroes in the end is nothing but the highest power of N in that product

Eg 21) Find the number of zeroes in 13! in base 10.

We need to find the highest power of 10 in 13! = highest power of 5 in 13!

This is because there will be lesser number of 5s present as compared to 2s (illustrated in the example above)


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Number of 2s = . total =6+3+1=10

Number of . number of zeroes will thus depend only on the number of 5s = 2

Eg 22) Find the number of zeroes in 25! In base 10.

We need to find the highest power of 10 in 25! Highest power of 10 will be the highest power of 5 in 25! As explained in theprevious example

. number of zeroes = 5+1=6

Eg 23) Find the number of zeroes in the end of 15! in base 12.

We need to find the highest power of 12 in 15! =highest power of 22 *3 in 15!

Highest power of 2 in . Total =11 .

Highest power of

Highest power of 3 in 15!= 5

Thus, the highest power of 12 in 15! = 5

Eg 24) Find the number of zeroes in the end of 200! In base 13

13 is a prime number, we need to find the highest power of 13 in 200!

. number of zeroes at the end of 200! In base 13 = 15+1=16

III) Number of factors of any factorial

There is a technique which can be used to find the number of factors in a factorial. Let us see how to do it with an example

Eg 25) Find the number of factors of 6!

STEP 1: Prime factorize 6! i.e. find out the highest power of all prime factors till 6 ( i.e. 2,3 and 5). 6! = 24*32*51

STEP2: Then use the formula

N=am*bn (a, b are the prime factors)

Then number of factors= (m+1)(n+1)

The number of factors= (4+1)(2+1)(1+1) =30

Answer=30

Eg 26) Find the number of factors of 12!

STEP 1: Prime factorize 12! i.e. find out the highest power of all prime factors till 12 ( i.e. 2,3,5,7,11). 12! = 210*35*52*7*11

STEP2: Then use the formula

N=am*bn (a, b are the prime factors)

Then number of factors= (m+1)(n+1)

The number of factors= (10+1)(5+1)(2+1)(1+1)(1+1) =792

Answer=792

Eg 27) The number of positive integers which divide (25)! are


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a) 213.33.52 b) 28.32.52 c) 211.32.5 d) 28.33.53

32! = 231 .314 . 57 .74 .112 ·13 2 ·17·19.23 .29.31

So it has (31+1) (14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1) = 213 .33 .52

Hence, choice (a) is the right answer

APPLICATION QUESTION BASED ON FACTORIAL

Eg 28) How many natural numbers are there such that their factorials are ending with 5 zeroes?

10! is 1*2*3*4*(5)*6*7*8*9*(2*5). From this we can see that highest power of 5 till 10! is 2( as there are 2 fives)

Continuing like this, 10!-14!, highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3).

Therefore, from 15! To 19! – The highest power of 5 will be 3.

20!-24! – Highest Power = 4

In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get highest power of 5 as 6. The answer to thequestion is therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

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