CAT Questions On Divisibility Rules Set-2

Understanding the divisibility properties of numbers holds significant importance in the CAT exam. Familiarity with divisibility rules is essential for solving various types of questions. If you haven’t acquired knowledge of these rules yet, you can conveniently access and download a comprehensive PDF that contains all the relevant formulas and concepts related to number systems. Going through this PDF will provide you with the necessary understanding of divisibility rules and enable you to approach related problems with confidence and accuracy. To access the PDF, please click on the provided link.

CAT Questions On Divisibility Rules

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Question 1: How many three digit numbers are divisible by 5 or 9?

a) 260
b) 280
c) 200
d) 180

Question 2: If 8A5146B is divisible by 88, then what is the value of AxB?

a) 4
b) 16
c) 8
d) 12
e)18

Question 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

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Question 4: The number A39K2 is completely divisible by both 8 and 11. Here both A and K are single digit natural numbers.
Which of the following is a possible value of A+K?

a) 8
b) 10
c) 12
d) 14

Question 5: 

A positive number is divided by 100 to get a remainder thrice as the quotient. If the number is divisible by 11, then how many such numbers are possible that are less than 100000?

Answer: 3

Solution:
Let the number be N and the quotient when divided by 100 be k. Then remainder is 3k. 3k < 100
N = 100k + 3k = 103k,
Also, N is divisible by 11.
=> k = 11p, where p is an integer.
=> N = 103*11p = 1133p
As N < 100000, it implies that p can range from 1 to [100000/1133] i.e between 1 and 88
=> So, p can range from 1 to 88.
Also 3k < 100 => 3 * 11p < 100 => p < 4
Hence, p can take values 1,2,3

Solutions: (1 to 4)

1) Answer (A)

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

2) Answer (d)

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, 14-A is divisible by 11.
Thus, the value of AxB = 4×3 = 12.

3) Answer (b)

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

4) Answer (b)
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 sum are
13, 5 and 10
Now 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 single digit 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 choice.

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