Table of contents  
Introduction  
Theorems of Divisibility  
Factors  
Factors Formulas  
Prime Factorisation  
Divisibility Rules  
Solved Questions 
A number 'x' is said to be completely divisible by another number ‘y’ if it leaves no remainder.
A factor of a number can be defined as a number that divides the given number without leaving any remainder.
Generally, factors based questions are of the following types:
(i) The number of factors.
(ii) The sum of factors.
(iii) Product of factors.
(iv) Odd and Even Factors.
(v) Number of Factors which are perfect squares for certain kinds of questions.
There are basically three types of formulas considered for factors. They are:
Let us assume N is a natural number, for which we need to find the factors. If we convert N into the product of prime numbers by prime factorization method, we can represent it as;
N = X^{a} × Y^{b} × Z^{c}
where X, Y and Z are the prime numbers and a, b and c are their respective powers.
The formula for the total number of factors for a given number is given by:
Total Number of Factors for N = (a+1) (b+1) (c+1) 
The formula for the sum of all factors is given by:
Sum of factors of N = [(X^{a+1}1)/X1] × [(Y^{b+1}1)/Y1] × [(Z^{c+1}1)/Z1] 
The formula for the product of all factors is given by:
Product of factors of N = N^{Total No. of Factors/2} 
We all know that every composite number can be written as a product of some prime numbers.
For example: We can write 90 as 2 × 3^{2} × 5. This process is called prime factorisation, and it is the first step to solve any questions related to factors.
Number of Factors of a Number
How to Find the Sum of Factors of a Number?
How to Find Number of Factors of a Number?
As the name suggests, divisibility rules or tests are the procedures used to check whether a number is divisible by another number without necessarily performing the actual division.
Divisibility Rules
The divisibility test for 1 does not have any condition for numbers. All numbers are divisible by 1 irrespective of how large the numbers are.
When any number is divided by 1, the result is the number itself.
Example: 5/1= 5 and 100000/1 = 100000.
A number is divisible by 2 if the last digit of the number is 2, 4, 6, 8 or 0.
Example: 102/2 = 51, 54/2 = 27, 66/2 = 33, 28/2 = 14 and 20/2 = 10
Example. Check if 288 is divisible by 2.
Ans: Given, 288 is a number.
If the last digit of 288 is divisible by 2, then 288 is also divisible by 2.
The last digit of 288 is 8, which is divisible by 2, such that;
8/2 = 4
Hence, 288 satisfy the divisibility rule for 2.
The divisibility test for 3 states that a number is completely divisible by 3 if the digits of the number are divisible by 3 or is a multiple of 3.
Divisibility Rule of 3
Example: Consider two numbers 308 and 207.
(i) Check 308 by summing its digits: 3 + 0 + 8 = 11, since 11 is not divisible by 3, then 308 is also not divisible by 3.
(ii) Check 207 by summing its digits: 2 + 0 + 7 = 9, since 9 is a multiple of 3, then 207 is also divisible by 3.
The divisibility test for 4 states that a number is divisible by 4 if the last two digits of the number are divisible by 4.
Divisibility Rule of 4
Example: Consider two numbers, 2508 and 2506.
(i) 2508 is divisible by 4 because the last two digits 08 are divisible by 4.
(ii) 2506 is not divisible by 4 because the last two digits 06 are not divisible by 4.
Q. Check if 195 is divisible by 4 or not.
Solution: As we can see, the last digit of 195 is 5, which is not divisible by 4.
Hence, 195 is not divisible by 4.
A number is divisible by 6 if its last digit is an even number or zero and the sum of the digits is a multiple of 3.
Example: 270 is divisible by 2 because the last digit is 0 and the sum of the digits is: 2 + 7 + 0 = 9 which is also divisible by 3.
Therefore, 270 is divisible by 6.
The divisibility test for 8 states that a number is divisible by 8 if its last three digits are divisible by 8.
Divisibility Rule of 8
The divisibility test for 9 is the same as the divisibility test for 3. If the sum of the digits of a number is divisible by 9, then the number is also divisible by 9.
Divisibility Rule of 9
Example: In a number like 78532, the sum of its digits is: 7+8+5+3+2 = 25.
This rule states that a number is divisible by 11 if the difference of the sum of alternative digits is divisible by 11.
Divisibility Rule of 11
Example: To check whether number 2143 is divisible by 11 or not.
The sum of alternative digits of each group: 2 + 4 = 6 and 1+ 3 = 4
Since 6  4 = 2, Therefore 2143 is not divisible by 11.
If the number is divisible by both 3 and 4, then the number is divisible by 12 exactly.
Divisibility Rule of 12
To check whether a number is divisible by 13, repeated addition of the last digit is done 4 times to the remaining number until a twodigit number is arrived at.
If the twodigit number is divisible by 13, then the whole number is also divisible by 13.
Divisibility Rule of 13
Example: 2795 → 279 + (5 x 4) → 279 + (20) → 299 → 29 + (9 x 4) → 29 + 36 →65.
Since the twodigit number is found to be 65, which is divisible by 13.
Therefore, the number 2795 is also divisible by 13.
Q.1. How many threedigit numbers are divisible by 5 or 9?
a) 260
b) 280
c) 200
d) 180
Ans: a) 260
Solution: Three digit numbers divisible by 5 or 9 = three digit numbers divisible by 5 + three digit numbers divisible by 9 – three digit numbers divisible by 5 and 9.
The three digit numbers divisible by 5 = 100, 105, 110….995
The sequence given is in A.P with common difference 5. Let 995 be the nth term of the A.P, then, 995 = 100 + (n – 1)5 = 100 + 5n – 5
Thus, n = 180 – (1)
The three digit numbers divisible by 9 = 108, 118, … 999
The sequence given is in A.P with common difference 9. Let 999 be the pth term of the A.P, then, 999 = 108 + (p – 1)9 = 108 + 9p – 9
Thus, p = 100 – (2)
The three digit numbers divisible by 45 = 135, 180, …990
The sequence given is in A.P with common difference 45. Let 990 be the qth term of the A.P, then
990 = 135 + (q – 1)45 = 135 + 45q – 45
Thus, q = 20 – (3)
Thus, from (1), (2) and (3) the three digit numbers divisible by 5 or 9 = 180 + 100 – 20 = 260 so, 260 three digit numbers are divisible by 5 or 9.
Q.2. If 8A5146B is divisible by 88, then what is the value of AxB?
a) 4
b) 16
c) 8
d) 12
e)18
Ans: d) 12
Solution: Since the given number is divisible by 8, the last three digits should also be divisible by 8. Only when B = 4, 46B is a multiple of 8. Thus, B = 4.
As the given number is divisible by 11, the difference between the sum of its odd digits and even digits must be a multiple of 11.
Thus, (8 + 5 + 4 + 4) – (A + 1 + 6) = 14 – A should be divisible by 11. Only when A = 3, 14A is divisible by 11.
Thus, the value of AxB = 4×3 = 12.
Q.3. What is the number of even factors of 36000 which are divisible by 9 but not by 36?
a) 20
b) 4
c) 10
d) 12
Ans: b) 4
Solution: 36000 = 2^{5} * 3^{2} * 5^{3}
Since we are talking of even factors, there must be at least one 2 in the required factors.
Since the number is divisible by 9, we must have both the threes.
We cannot have more than 1 two as it will make the number divisible by 36.
So we have 1 way of choosing 2, 1 way of choosing 3, 4 ways of choosing 5.
Thus the required number of factors are
1*1*4 = 4
Q.4. The number A39K2 is completely divisible by both 8 and 11. Here both A and K are singledigit natural numbers. Which is a possible value of A+K?
a) 8
b) 10
c) 12
d) 14
Ans: b) 10
Solution: The number is divisible by 11, so the difference between the sum of the digits at the odd places and the digits at the even places is either 0 or a multiple of 11.
Let the difference be a 0, so
11 + A = 3 + K
=> K – A = 8, the only possible value is 9,1
Now we have to check if it satisfies the divisibility by 8 test.
K= 9 makes the last 3 digits 992. This is divisible by 8.
Let’s check for other cases when the difference is 11
11 + A – 3 – K = 11 => A – K = 3
The possible values in this case are (9,6), ( 8,5), (7,4), (6,3), (5,2), (4,1)
Among these cases, only (8,5) and (4,1) will be divisible by 8. So the possible values of the sum are 13, 5, and 10.
Now, the difference between the sum of odd and even places cannot be 22
11 + A – 3 – K = 22 => A – K = 14
Since, both A and K are singledigit natural numbers, this is not possible.
Thus the only possible values of sum are 5, 10, and 13.
In the given options only 10 is there. So it is the correct answer.
175 videos131 docs110 tests

1. What are some common divisibility rules tested in the CAT exam? 
2. How can prime factorization be used to find factors of a number in the CAT exam? 
3. How can factor formulas be applied to solve CAT questions on divisibility? 
4. What is the significance of understanding divisibility rules in the CAT exam? 
5. How can the theorems of divisibility be applied to solve CAT questions related to factors? 

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