Factorials And Their Applications

With the approaching date of CAT exam, the candidates are required to speed up their preparation and get well versed with all the important topics of the exam. As the CAT exam is extremely competitive, the candidates are required to be well prepared with all the CAT important topics to ace the exam.

One of the most important CAT exam topics is the Factorial. A thorough understanding of Factorials is important because they play a pivotal role in the Quantitative Aptitude topics like Concepts surrounding Numbers, Permutation and Combination, etc. So, here is an introductory lesson on the Factorials and their applications.

Factorial Formula:

The factorial of a number can be defined as:

 

 n! = 1 × 2 × 3 ×…(n-1) × n

 

Factorial of a Number: Examples

The factorial of 5 is:

5!= 1 × 2 × 3 × 4 × 5=120.

The factorial of 3 is:

3!=1 × 2 × 3=6.

Application of Factorials:

The applications of Factorials have been categorized into the following

  • The highest power of a number in a factorial or in a product
  • Number of zeros in the end of a factorial or a product
  • Number of factors of any factorial
  • The highest power of a number in a factorial or in a product

Questions on Highest Power in a Factorial:

Questions based on highest power in a factorial are seen often in the CAT paper. Here is an explanation on the topic in detail.

 

What is the 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 on this can be categorized based on the nature of the number whose highest power needs to be found in the factorial, i.e.

1) The 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.

 

2) Find the highest power of 5 in 100!

Solution: 100/5=20; 20/5=4;

Adding the quotients, its 20+4=24.

So highest power of 5 in 100! = 24

 

Alternative Method :

100/5+100/52 =20+4=24 (up to 52 is taken as it is the highest power of 5 which is less than 100).

3) 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.

Illustrations On Factorials:

1) To find the highest power of 10 in 100!

Solution:

Factorize 10=5 × 2.

1.The 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 take the lesser value i.e. 24 is the answer.

 

2) Highest power of 12 in 100!

Solution:

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:

100/2= 50; 50/2= 25; 25/2= 12; 12/2= 6;6/2= 3; 3/2= 1

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

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

Now for the highest power of 3,

100/3= 33; 33/3= 11; 11/3= 3; 3/3= 1

So, highest power of 3 = 48

Also, Highest power of 12 = 48

 

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

Answer:

In base 10, the 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 previous example, you will get a lesser number of 5s than 2s.

that is why one needs to calculate only the number of 5s) Similarly, for base 12, the number of zeros will depend on the number 12 itself.

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

 

4) Find the number of zeroes in 13! In base 10?

Answer:

The highest power of 10 in 13! is to be found i.e. = highest power of 5 in 13!

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

Number of 2s= 13/2=6; 6/2=3; 3/2=1.

total =6+3+1=10 Number of 5s = 13/5 = 2. number of zeroes will thus depend only on the number of 5s = 2

 

5) Find the number of zeroes in 25! In base 10

Solution:

The highest power of 10 in 25! is to be found.

Highest power of 10 will be the highest power of 5 in 25! As explained in the previous example 25/5=5; 5/5=1.

number of zeroes = 5+1=6

 

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

Solution:

One needs to find the highest power of 12 in 15! =highest power of 22 × 3 in 15!

Highest power of 2 in 15!= 15/2=7; 7/2=3; 3/2=1. Total =11.

Highest power of 22 = 11/2=5

Highest power of 3 in 15!= 5

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

 

7) Find the highest power of 13 in 200!

Solution

13 is a prime number and so, one needs to find the highest power of 13 in 200! which is 200/13=15;15/13=1.

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

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:

 

8) Find the number of factors of 6!

Solution:

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

STEP 2: Then use the formula N=a× 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

So, answer=30

 

9) Find the number of factors of 12!

Solution:

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=a× b(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.

So, answer=792

 

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

a) 213.33.52

b) 28.32.52

c) 211.32.5

d) 28.33.53

Solution:

option (a)

32! = 231 × 314 × 5× 7× 11× 132 × 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 Based Questions On Factorials:

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

Solution:

10! is 1× 2× 3× 4× (5) × 6 × 7 × 8 × 9 × (2 × 5).

From this it can be seen that highest power of 5 till 10! is 2( as there are 2 fives) Continuing like this, 10!-14!, the 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,

One extra five is present, as 25=5 × 5. Therefore, 25! to 29!, one will get the highest power of 5 as 6.

The answer to the question is, therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

The candidates are suggested to go through the lesson properly to strengthen the fundamentals of factorials and its application. It is also recommended to visit CAT Syllabus to get well acquainted with the CAT topics and prepare accordingly.

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