Download the free PDF of RD Sharma Solutions Class 8 Maths Chapter 5 Playing with Numbers. Experts at BYJU’S have designed these solutions in a very clear and precise manner that helps students solve problems in the most efficient possible ways. Securing good marks has become much easier with the usage of Solutions of RD Sharma Class 8 Maths. In Exercise 5.2, we will discuss problems based on the test of divisibility by 10, 5, 2, 9, 3, 6, 11, and 4. Students can refer to and download RD Sharma Class 8 Solutions for Maths from the links provided below.
RD Sharma Solutions for Class 8 Maths Chapter 5 Playing with Numbers Exercise 5.2
Access Answers to Maths RD Sharma Solutions for Class 8 Exercise 5.2 Chapter 5 – Playing with Numbers
1. Given that the number
is divisible by 3, where a is a digit, what are the possible values of a?
Solution:
We know that the given number
is divisible by 3.
And, if a number is divisible by 3, then the sum of digits must be a multiple of 3.
i.e., 3 + 5 + a + 6 + 4 = multiple of 3
a + 18 = 0, 3, 6, 9, 12, 15…..
Here ‘a’ is a digit, where ‘a’ can have values between 0 and 9.
a + 18 = 18 which gives a = 0.
a + 18 = 21 which gives a = 3.
a + 18 = 24 which gives a = 6.
a + 18 = 27 which gives a = 9.
∴ a = 0, 3, 6, 9
2. If x is a digit such that the number 
is divisible by 3, find possible values of x.
Solution:
We know that the given number
is divisible by 3.
And, if a number is divisible by 3, then the sum of digits must be a multiple of 3.
i.e., 1 + 8+ x + 7 + 1 = multiple of 3
x + 17 = 0, 3, 6, 9, 12, 15…..
Here ‘x’ is a digit, where ‘x’ can have values between 0 and 9.
x + 17 = 18 which gives x = 1.
x + 17 = 21 which gives x = 4.
x + 17 = 24 which gives x = 7.
∴ x = 1, 4, 7
3. If x is a digit of the number 
such that it is divisible by 9, find possible values of x.
Solution:
We know that the given number
is divisible by 9.
And, if a number is divisible by 9, then the sum of digits must be a multiple of 9.
i.e., 6 + 6 + 7 + 8 + 4 + x = multiple of 9
x + 31 = 0, 9, 18, 27…..
Here ‘x’ is a digit, where ‘x’ can have values between 0 and 9.
x + 31 = 36 which gives x = 5.
∴ x = 5
4. Given that the number 
is divisible by 9, where y is a digit, what are the possible values of y?
Solution:
We know that the given number
is divisible by 9.
And, if a number is divisible by 9, then the sum of digits must be a multiple of 9.
i.e., 6 + 7+ y + 1 + 9 = multiple of 9
y + 23 = 0, 9, 18, 27…..
Here ‘y’ is a digit, where ‘y’ can have values between 0 and 9.
y + 23 = 27 which gives y = 4.
∴ y = 4
5. If 
is a multiple of 11, where x is a digit, what is the value of x?
Solution:
We know that the given number
is a multiple of 11.
A number is divisible by 11 if and only if the difference between the sum of odd and even place digits is a multiple of 11.
i.e., Sum of even placed digits – Sum of odd placed digits = 0, 11, 22…
x – (3+2) = 0, 11, 22…
x – 5 is a multiple of 11
x – 5 = 0
∴ x = 5
6. If 
is a number with x as its tens digit such that it is divisible by 4. Find all possible values of x.
Solution:
We know that the given number
is divisible by 4.
A number is divisible by 4 only when the number formed by its digits in unit’s and ten’s place is divisible by 4.
i.e., x2 is divisible by 4
Expanding x2,
10x + 2 = multiple of 4
Here x2 can take values 2, 12, 22, 32, 42, 52, 62, 72, 82, 92
So values 12, 32, 52, 72 and 92 are divisible by 4.
∴ x can take values 1, 3, 5, 7 and 9.
7. If x denotes the digit at the hundreds place of the number 
such that the number is divisible by 11. Find all possible values of x.
Solution:
We know that the given number
is divisible by 11.
A number is divisible by 11 if and only if the difference between the sum of odd and even place digits is a multiple of 11.
i.e., Sum of even placed digits – Sum of odd placed digits = 0, 11, 22…
(6 + x + 9) – (7+1) = 0, 11, 22…
x + 7 is a multiple of 11
x + 7 = 11
∴ x = 4
8. Find the remainder when 981547 is divided by 5. Do that without doing actual division.
Solution:
We know that if a number is divided by 5, then the remainder is obtained by dividing just the unit place by 5.
i.e., 7 ÷ 5 gives 2 as a remainder.
∴ The remainder will be 2 when 981547 is divided by 5.
9. Find the remainder when 51439786 is divided by 3. Do that without performing actual division.
Solution:
We know that if a number is divided by 3, then the remainder is obtained by dividing the sum of digits by 3.
Here, the sum of digits (5+1+4+3+9+7+8+6) is 43.
i.e., 43 ÷ 3 gives 1 as a remainder.
∴ The remainder will be 1 when 51439786 is divided by 3.
10. Find the remainder, without performing actual division, when 798 is divided by 11.
Solution:
We know that if a number is divided by 11, then the remainder is the difference between the sum of even and odd digit places.
i.e., Remainder = 7 + 8 – 9 = 6
∴ The remainder will be 6 when 798 is divided by 11.
11. Without performing actual division, find the remainder when 928174653 is divided by 11.
Solution:
We know that if a number is divided by 11, then the remainder is the difference between the sum of even and odd digit places.
i.e., Remainder = 9 + 8 + 7 + 6 + 3 – 2 – 1 – 4 – 5 = 33 – 12 = 21
∴ 21 ÷ 11 gives 10 as the remainder
∴ The remainder will be 10 when 928174653 is divided by 11.
12. Given an example of a number which is divisible by
(i) 2 but not by 4.
(ii) 3 but not by 6.
(iii) 4 but not by 8.
(iv) Both 4 and 8 but not by 32
Solution:
(i) 2 but not by 4.
Any number which follows the formula of 4n + 2 is an example of a number divisible by 2 but not by 4.
i.e., 10 where n = 1.
(ii) 3 but not by 6.
Any number which follows the formula of 6n + 3 is an example of a number divisible by 3 but not by 6.
i.e., 15 where n = 1.
(iii) 4 but not by 8.
Any number which follows the formula of 8n + 4 is an example of a number divisible by 4 but not by 8.
i.e., 28 where n = 1.
(iv) Both 4 and 8 but not by 32
Any number which follows the formula of 32n + 8 or 32n + 16 or 32n +24 is an example of a number divisible by both 4 and 8 but not by 32.
i.e., 48 where n = 1.
13. Which of the following statements is true?
(i) If a number is divisible by 3, it must be divisible by 9.
(ii) If a number is divisible by 9, it must be divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
(iv) If a number is divisible by 8, it must be divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
(ix) If two numbers are co-prime, at least one of them must be a prime number.
(x) The sum of two consecutive odd numbers is always divisible by 4.
Solution:
(i) If a number is divisible by 3, it must be divisible by 9.
False
Because any number which follows the formula 9n + 3 or 9n + 6 violates the statement.
For example, 6, 12…
(ii) If a number is divisible by 9, it must be divisible by 3.
True
Because 9 is a multiple of 3, any number divisible by 9 is also divisible by 3.
(iii) If a number is divisible by 4, it must be divisible by 8.
False
Because any number which follows the formula 8n + 4 violates the statement.
For example, 4, 12, 20….
(iv) If a number is divisible by 8, it must be divisible by 4.
True
Because 8 is a multiple of 4, any number divisible by 8 is also divisible by 4.
(v) A number is divisible by 18, if it is divisible by both 3 and 6.
False
Because for example, 24 is divisible by both 3 and 6 but not divisible by 18.
(vi) If a number is divisible by both 9 and 10, it must be divisible by 90.
True
Because 90 is the GCD of 9 and 10, any number divisible by both 9 and 10 is also divisible by 90.
(vii) If a number exactly divides the sum of two numbers, it must exactly divide the numbers separately.
False
Because let us consider an example 6 divided 30, but 6 divides none of 13 and 17 as both are prime numbers.
(viii) If a number divides three numbers exactly, it must divide their sum exactly.
True
Because if x, y and z are three numbers, each of x, y and z are divided by a number (say q), then (x + y + z) is also divisible by q.
(ix) If two numbers are co-prime, at least one of them must be a prime number.
False
Because 16 and 21 are co-prime, but none of them is prime.
(x) The sum of two consecutive odd numbers is always divisible by 4.
True
Because 3+5 = 8, which is divisible by 4.
