RD Sharma Solutions for Class 6 Maths Chapter 2 Playing with Numbers

RD Sharma Solutions for Class 6 Maths Chapter 2: Playing with Numbers are expertly crafted by mathematics professionals with deep conceptual knowledge and teaching experience. These chapter-wise solutions are designed as per the latest CBSE syllabus, ensuring accuracy and relevance for school exams. 

The RD Sharma Solutions for Class 6 Maths Chapter 2 solutions PDF acts as a helpful reference tool for students, improving their problem-solving skills and boosting confidence during preparation. Each problem is solved in a clear, step-by-step manner, making the concepts easy to understand for young learners. 

Whether used for revision or practice, these RD Sharma Class 6 Maths solutions help students score better in exams by strengthening their understanding of topics like divisibility, prime numbers, LCM, HCF, and more.

RD Sharma Solutions for Class 6 Maths Chapter 2 Playing with Numbers 

For students looking to master the concepts of Playing with Numbers, the RD Sharma Solutions for Class 6 Chapter 2 are an excellent resource. These solutions are available in PDF format, making them ideal for offline study and easy access anytime. Whether you're aiming to strengthen your basics or revise for your upcoming exams, the RD Sharma Class 6 Maths PDF download offers chapter-wise clarity on important topics like factors, multiples, prime numbers, LCM, and HCF.

Learners preparing for the academic year 2025–26 can benefit greatly from using the RD Sharma Class 6 full book PDF with solutions, which aligns with the latest CBSE syllabus. For additional self-practice, students can also download Class 6 Maths Chapter 2 worksheets with answers in PDF, which are excellent for self-assessment and exam revision.

These solutions also serve as a great support tool alongside NCERT Solutions for Class 6 Maths, making them a comprehensive choice for CBSE students. With step-by-step explanations and concept-based approaches, RD Sharma Solutions for Chapter 2 help build strong foundational skills in mathematics.

RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing with Numbers PDF

Chapter 2 – Playing with Numbers is a crucial part of Class 6 Maths, and RD Sharma explains it in a clear and student-friendly way. This chapter helps students dive deeper into how numbers work, focusing on important topics like factors, multiples, prime and composite numbers, divisibility rules, LCM, and HCF. Having access to the RD Sharma Class 6 Chapter 2 PDF with step-by-step solutions allows students to practice regularly, clear doubts instantly, and build strong problem-solving skills from the ground up.

The solutions are organized exercise-wise, making it easier for students to learn one concept at a time and track their progress. These well-structured solutions follow the latest CBSE curriculum, making them perfect for revision, homework, or exam prep. Each question is explained in a simple and detailed manner to help students fully understand the logic behind every answer.

Chapter 2 – Playing with Numbers Exercises:

• Exercise 2.1 – Factors and Multiples
• Exercise 2.2 – Prime and Composite Numbers
• Exercise 2.3 – Tests of Divisibility
• Exercise 2.4 – Common Factors and Common Multiples
• Exercise 2.5 – LCM and HCF
• Exercise 2.6 – Word Problems and Applications

Access RD Sharma Class 6 Maths Chapter 2 Solutions – Playing with Numbers

Exercise 2.1 page: 2.5

1. Define:

(i) factor

(ii) multiple

Give four examples each.

Solution:

(i) A factor of a number is an exact divisor of that number.

Example:

1. 5 and 2 are factors of 10 i.e. 5 × 2 = 10
2. 4 and 3 are factors of 12 i.e. 4 × 3 = 12
3. 4 and 2 are factors of 8 i.e. 4 × 2 = 8
4. 2 and 6 are factors of 12 i.e. 2 × 6 = 12

(ii) A multiple of a number is a number obtained by multiplying it by a natural number.

Example:

1. 10 is a multiple of 2 i.e. 2 × 5 = 10
2. 12 is a multiple of 4 i.e. 4 × 3 = 12
3. 8 is a multiple of 2 i.e. 2 × 4 = 8
4. 21 is a multiple of 3 i.e. 3 × 7 = 21

2 - Write all factors of each of the following numbers:

(i) 60

(ii) 76

(iii) 125

(iv) 729

Solution:

(i) 60

It can be written as

1 × 60 = 60

2 × 30 = 60

3 × 20 = 60

4 × 15 = 60

5 × 12 = 60

6 × 10 = 60

Therefore, the factors of 60 are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60.

(ii) 76

It can be written as

1 × 76 = 76

2 × 38 = 76

4 × 19 = 76

Therefore, the factors of 76 are 1, 2, 4, 19, 38 and 76.

(iii) 125

It can be written as

1 × 125 = 125

5 × 25 = 125

Therefore, the factors of 125 are 1, 5, 25 and 125.

(iv) 729

It can be written as

1 × 729 = 729

3 × 243 = 729

9 × 81 = 729

27 × 27 = 729

Therefore, the factors of 729 are 1, 3, 9, 27, 81, 243 and 729.

3. Write first five multiples of each of the following numbers:

(i) 25

(ii) 35

(iii) 45

(iv) 40

Solution:

(i) 25

It can be written as

1 × 25 = 25

2 × 25 = 50

3 × 25 = 75

4 × 25 = 100

5 × 25 = 125

Therefore, the first five multiples of 25 are 25, 50, 75, 100 and 125.

(ii) 35

It can be written as

1 × 35 = 35

2 × 35 = 70

3 × 35 = 105

4 × 35 = 140

5 × 35 = 175

Therefore, the first five multiples of 35 are 35, 70, 105, 140 and 175.

(iii) 45

It can be written as

1 × 45 = 45

2 × 45 = 90

3 × 45 = 135

4 × 45 = 180

5 × 45 = 225

Therefore, the first five multiples of 45 are 45, 90, 135, 180 and 225.

(iv) 40

It can be written as

1 × 40 = 40

2 × 40 = 80

3 × 40 = 120

4 × 40 = 160

5 × 40 = 200

Therefore, the first five multiples of 40 are 40, 80, 120, 160 and 200.

4. Which of the following numbers have 15 as their factor?

(i) 15625

(ii) 123015

Solution:

(i) 15625

We know that 15 is not a factor of 15625 as it is not a divisor of 15625.

(ii) 123015

We know that 15 is a factor of 123015 as it is a divisor of 123015 because 8201 × 15 = 123015.

5. Which of the following numbers are divisible by 21?

(i) 21063

(ii) 20163

Solution:

(i) 21063

We know that the sum of digits = 2 + 1 + 0 + 6 + 3 = 12 which is divisible by 3

So 21063 is divisible by 3

A number is divisible by 7 if the difference between two times ones digit and the number formed by other digit is 0 or multiple of 7.

We get

2106 – (2 × 3) = 2100 which is a multiple of 7.

Therefore, 21063 is divisible by 21.

(ii) 20163

We know that the sum of digits = 2 + 0 + 1 + 6 + 3 = 12 which is divisible by 3

So 20163 is divisible by 3

A number is divisible by 7 if the difference between two times ones digit and the number formed by other digit is 0 or multiple of 7.

We get

2016 – (2 × 3) = 2010 which is not a multiple of 7.

Therefore, 20163 is not divisible by 21.

6. Without actual division show that 11 is a factor of each of the following numbers:

(i) 1111

(ii) 11011

(iii) 110011

(iv) 1100011

Solution:

(i) 1111

Sum of digits at the odd places = 1 + 1 = 2

Sum of digits at the even places = 1 + 1 = 2

So the difference between the two sums = 2 – 2 = 0

Hence, 1111 is divisible by 11 as the difference between the two sums is zero.

(ii) 11011

Sum of digits at the odd places = 1 + 0 + 1 = 2

Sum of digits at the even places = 1 + 1 = 2

So the difference between the two sums = 2 – 2 = 0

Hence, 11011 is divisible by 11 as the difference between the two sums is zero.

(iii) 110011

Sum of digits at the odd places = 1 + 0 + 1 = 2

Sum of digits at the even places = 1 + 0 + 1 = 2

So the difference between the two sums = 2 – 2 = 0

Hence, 110011 is divisible by 11 as the difference between the two sums is zero.

(iv) 1100011

Sum of digits at the odd places = 1 + 0 + 0 + 1 = 2

Sum of digits at the even places = 1 + 0 + 1 = 2

So the difference between the two sums = 2 – 2 = 0

Hence, 1100011 is divisible by 11 as the difference between the two sums is zero.

7. Without actual division show that each of the following numbers is divisible by 5:

(i) 55

(ii) 555

(iii) 5555

(iv) 50005

Solution:

(i) 55

The units digit in 55 is 5.

Therefore, 55 is divisible by 5.

(ii) 555

The units digit in 555 is 5.

Therefore, 555 is divisible by 5.

(iii) 5555

The units digit in 5555 is 5.

Therefore, 5555 is divisible by 5.

(iv) 50005

The units digit in 50005 is 5.

Therefore, 50005 is divisible by 5.

8. Is there any natural number having no factor at all?

Solution:

No. All the natural numbers are a factor of itself.

9. Find numbers between 1 and 100 having exactly three factors.

Solution:

We know that the numbers between 1 and 100 which have exactly three factors are 4, 9, 25 and 49.

Factors of 4 are 1, 2 and 4.

Factors of 9 are 1, 3 and 9.

Factors of 25 are 1, 5 and 25.

Factors of 49 are 1, 7 and 49.

10. Sort out even and odd numbers:

(i) 42

(ii) 89

(iii) 144

(iv) 321

Solution:

We know that

The numbers which are divisible by 2 are even and those which are not divisible by 2 are odd numbers.

So we get

42 and 144 are even numbers and 89 and 321 are odd numbers.

Exercise 2.2 page: 2.7

1. Find the common factors of:

(i) 15 and 25

(ii) 35 and 50

(iii) 20 and 28

Solution:

(i) 15 and 25

We know that

1 × 15 = 15

3 × 5 = 15

Factors of 15 are 1, 3, 5 and 15.

We know that

1 × 25 = 25

5 × 5 = 25

Factors of 25 are 1, 5 and 25.

Therefore, the common factors of 15 and 25 are 1 and 5.

(ii) 35 and 50

We know that

1 × 35 = 35

5 × 7 = 35

Factors of 35 are 1, 5, 7 and 35.

We know that

1 × 50 = 50

2 × 25 = 50

5 × 10 = 50

Factors of 50 are 1, 2, 5, 10, 25 and 50.

Therefore, the common factors of 35 and 50 are 1 and 5.

(iii) 20 and 28

We know that

1 × 20 = 20

2 × 10 = 20

4 × 5 = 20

Factors of 20 are 1, 2, 4, 5, 10 and 20.

We know that

1 × 28 = 28

2 × 14 = 28

7 × 4 = 28

Factors of 28 are 1, 2, 4, 7, 14 and 28.

Therefore, the common factors of 20 and 28 are 1, 2 and 4.

2. Find the common factors of:

(i) 5, 15 and 25

(ii) 2, 6 and 8

Solution:

(i) 5, 15 and 25

We know that

For 5

1 × 5 = 5

Factors of 5 are 1 and 5

For 15

1 × 15 = 15

3 × 5 = 15

Factors of 15 are 1, 3, 5 and 15

For 25

1 × 25 = 25

5 × 5 = 25

Factors of 25 are 1, 5 and 25

Therefore, the common factors of 5, 15 and 25 are 1 and 5.

(ii) 2, 6 and 8

We know that

For 2

1 × 2 = 2

Factors of 2 are 1 and 2

For 6

1 × 6 = 6

2 × 3 = 6

Factors of 6 are 1, 2, 3 and 6

For 8

1 × 8 = 8

2 × 4 = 8

Factors of 8 are 1, 2, 4 and 8

Therefore, the common factors of 2, 6 and 8 are 1 and 2.

3. Find first three common multiples of 6 and 8.

Solution:

We know that the multiples of 6 are

6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72 …..

Multiples of 8 are

8, 16, 24, 32, 40, 48, 56, 64, 72, 80 ……

Hence, the first three common multiples of 6 and 8 are 24, 48 and 72.

4. Find first two common multiples of 12 and 18.

Solution:

We know that multiples of 12 are

12, 24, 36, 48, 60, 72, 84, 96, 108, 120 …..

Multiples of 18 are

18, 36, 54, 72, 90, 108, 126, 144, 162, 180 ….

Hence, the first two common multiples of 12 and 18 are 36 and 72.

5. A number is divisible by both 7 and 16. By which other number will that number be always divisible?

Solution:

It is given that a number is divisible by both 7 and 16.

We know that the factors of 7 are 1 and 7 and the factors of 16 are 1, 2, 4, 8 and 16

Hence, the common factor of 7 and 16 is 1 and the number is divisible by 1.

6. A number is divisible by 24. By what other numbers will that number be divisible?

Solution:

It is given that a number is divisible by 24

We know that the factors of 24 are

1, 2, 3, 4, 6, 8, 12 and 24

Therefore, the number is divisible by 1, 2, 3, 4, 6, 8 and 12.

Exercise 2.3 page: 2.10

1. What are prime numbers? List all primes between 1 and 30.

Solution:

A number is called a prime number if it has no factor other than 1 and the number itself.

For example 2, 3, 5, 7, 11 and 13 are prime numbers.

The list of primes between 1 and 30 are

2, 3, 5, 7, 11, 13, 17, 19, 23 and 29.

2. Write all prime numbers between:

(i) 10 and 50

(ii) 70 and 90

(iii) 40 and 85

(iv) 60 and 100

Solution:

(i) The prime numbers between 10 and 50 are

11, 13, 17, 19, 23, 29, 31, 37, 41, 43 and 47.

(ii) The prime numbers between 70 and 90 are

71, 73, 79, 83 and 89.

(iii) The prime numbers between 40 and 85 are

41, 43, 47, 53, 59, 61, 67, 71, 73, 79 and 83.

(iv) The prime numbers between 60 and 100 are

61, 67, 71, 73, 79, 83, 89 and 97.

3. What is the smallest prime number? Is it an even number?

Solution:

2 is the smallest prime number.

We know that 2 is an even prime number since it is divisible by 2.

4. What is the smallest odd prime? Is every odd number a prime number? If not, give an example of an odd number which is not prime.

Solution:

3 is the smallest odd prime number.

No, every odd number is not a prime number.

Example: 9 is an odd number having factors 1, 3 and 9 and is not a prime number.

5. What are composite numbers? Can a composite number be odd? If yes, write the smallest odd composite number.

Solution:

A number is called a composite number if it has at least one factor other than 1 and the number itself.

Example: 4, 6, 8, 9, 10 and 15 are composite numbers.

Yes, a composite number can be odd and 9 is the smallest odd number.

6. What are twin-primes? Write all pairs of twin-primes between 50 and 100.

Solution:

Two prime numbers are known as twin-primes if there is only one composite number between them.

Example: 3, 5; 5, 7; 11, 13; 17, 19; 29, 31; 41, 43; 59, 61 and 71, 73 are the pairs of twin-primes between 1 and 100.

The pairs of twin-primes between 50 and 100 are

59, 61 and 71, 73.

7. What are co-primes? Give examples of five pairs of co-primes. Are co-primes always prime? If no, illustrate your answer by an example.

Solution:

Two numbers are said to be co-prime if they do not have a common factor other than 1.

2, 3; 3, 4; 4, 5; 5, 6; 6, 7 are the examples of five pairs of co-primes.

No. 15, 16 are co-prime but both of the numbers are not prime.

8. Which of the following pairs are always co-prime?

(i) two prime numbers

(ii) one prime and one composite number

(iii) two composite numbers

Solution:

(i) Two prime numbers are always co-prime to each other.

For example: The numbers 7 and 11 are co-prime to each other.

(ii) One prime and one composite number are not always co-prime.

For example: The numbers 3 and 21 are not co-prime to each other.

(iii) Two composite numbers are not always co-prime to each other.

For example: 4 and 6 are not co-prime to each other.

9. Express each of the following as a sum of two or more primes:

(i) 13

(ii) 130

(iii) 180

Solution:

(i) We know that

The number 13 can be written as

13 = 3 + 3 + 7 which is the sum of two or more primes.

(ii) We know that

The number 130 can be written as

130 = 59 + 71 which is the sum of two or more primes.

(iii) We know that

The number 180 can be written as

180 = 79 + 101 which is the sum of two or more primes.

10. Express each of the following numbers as the sum of two odd primes:

(i) 36

(ii) 42

(iii) 84

Solution:

(i) We know that

The number 36 can be written as

36 = 7 + 29 which is the sum of two odd primes

(ii) We know that

The number 42 can be written as

42 = 5 + 37 which is the sum of two odd primes

(iii) We know that

The number 84 can be written as

84 = 17 + 67 which is the sum of two odd primes

11. Express each of the following numbers as the sum of three odd prime numbers:

(i) 31

(ii) 35

(iii) 49

Solution:

(i) We know that

The number 31 can be written as

31 = 5 + 7 + 19 which is the sum of three odd prime numbers

(ii) We know that

The number 35 can be written as

35 = 5 + 7 + 23 which is the sum of three odd prime numbers

(iii) We know that

The number 49 can be written as

49 = 3 + 5 + 41 which is the sum of three odd prime numbers

12. Express each of the following numbers as the sum of twin primes:

(i) 36

(ii) 84

(iii) 120

Solution:

(i) We know that

The number 36 can be written as

36 = 17 + 19 which is the sum of twin primes

(ii) We know that

The number 84 can be written as

84 = 41 + 43 which is the sum of twin primes

(iii) We know that

The number 120 can be written as

120 = 59 + 61 which is the sum of twin primes

13. Find the possible missing twins for the following numbers so that they become twin primes:

(i) 29

(ii) 89

(iii) 101

Solution:

(i) We know that 27 and 31 are the possible missing twins for 29.

31 is a prime number and 27 is not a prime number

Therefore, 31 is the missing twin.

(ii) We know that 87 and 91 are the possible missing twins for 89

Both the numbers are not prime.

Therefore, 89 has no twin.

(iii) We know that 99 and 103 are the possible missing twins for 101

103 is a prime number and 99 is not a prime number

Therefore, 103 is the missing twin.

14. A list consists of the following pairs of numbers:

(i) 51, 53; 55, 57; 59, 61; 63, 65; 67, 69; 71, 73

Categorize them as pairs of

(i) co-primes

(ii) primes

(iii) composites

Solution:

(i) Two numbers are said to be co-prime if they do not have a common factor other than 1.

Therefore, all the given pairs of numbers are co-primes.

(ii) A number is called a prime number if it has no factor other than 1 and the number itself.

Therefore, 59, 61; 71, 73 are pairs of prime numbers.

(iii) A number is called a composite number if it has at least one factor other than 1 and the number itself.

Therefore, 55, 57; 63, 65 are pairs of composite numbers.

15. For a number, greater than 10, to be prime what may be the possible digit in the units place?

Solution:

For a number, greater than 10, the possible digit in the units place to be prime can be 1, 3, 7 or 9.

For example: 11, 13, 17 and 19 are prime numbers which are greater than 10.

16. Write seven consecutive numbers less than 100 so that there is no prime number between them.

Solution:

The seven consecutive composite numbers less than 100 so that there is no prime number between them are

90, 91, 92, 93, 94, 95 and 96.

17. State true (T) or false (F):

(i) The sum of primes cannot be a prime.

(ii) The product of primes cannot be a prime.

(iii) An even number is composite.

(iv) Two consecutive numbers cannot be both primes.

(v) Odd numbers cannot be composite.

(vi) Odd numbers cannot be written as sum of primes.

(vii) A number and its successor are always co-primes.

Solution:

(i) False. 2 + 3 = 5 is a prime number.

(ii) True. The product of primes is a composite number.

(iii) False. Even number 2 is not composite.

(iv) False. The numbers 2 and 3 are consecutive and prime numbers.

(v) False. 9 is an odd number and is composite having factors 1, 3 and 9.

(vi) False. 9 is an odd number and the sum of prime number 7 + 2 = 9.

(vii) True. A number and its successor have only one common factor.

18. Fill in the blanks in the following:

(i) A number having only two factors is called a ……….

(ii) A number having more than two factors is called a …………

(iii) 1 is neither ………. Nor ………

(iv) The smallest prime number is ………..

(v) The smallest composite number is ……….

Solution:

(i) A number having only two factors is called a prime number.

(ii) A number having more than two factors is called a composite number.

(iii) 1 is neither prime nor composite.

(iv) The smallest prime number is 2.

(v) The smallest composite number is 4.

Benefits of Using RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing with Numbers

The chapter Playing with Numbers lays the foundation for understanding factors, multiples, divisibility, and number properties. Using RD Sharma Class 6 Chapter 2 Solutions helps students grasp these core concepts effectively through clear, structured explanations.

Key Benefits:

1. Helps students master the concepts of factors, multiples, LCM, and HCF with clarity.
2. Boosts problem-solving speed and accuracy with step-by-step solved examples.
3. Improves understanding of divisibility rules through easy-to-follow methods.
4. Offers CBSE Class 6 Maths aligned questions for exam-oriented practice.
5. Encourages independent learning with self-paced exercises in PDF format.
6. Builds confidence to solve word problems and logic-based numerical questions.
7. Supports revision and homework with well-structured answer explanations.

Attributes of RD Sharma Solutions for Class 6 Maths Chapter 2 – Playing with Numbers

RD Sharma’s solutions stand out due to their structured approach, exam alignment, and ease of understanding. These qualities make the solutions an essential companion for Class 6 Maths learners.

Notable Features:

• Designed by expert Mathematics educators with CBSE curriculum knowledge.
• Provides complete coverage of all exercises from Chapter 2 in logical sequence.
• Solutions are written in student-friendly language for easy comprehension.
• Available in downloadable PDF format for offline practice anytime.
• Each topic is supported with illustrations, shortcuts, and explanations.
• Focuses on developing analytical and critical thinking skills through Maths.
• Includes chapter-wise and concept-wise breakdowns to avoid confusion.

Why Should Students Choose RD Sharma Solutions for Class 6 Chapter 2 – Playing with Numbers?

Choosing the right study material can make a big difference in understanding Mathematics. RD Sharma Solutions provide a solid foundation in numbers while preparing students for higher-level concepts in later classes.

1. Covers every topic in Chapter 2 from basic concepts to advanced applications.
2. Supports students preparing for school exams and entrance-level tests.
3. Encourages conceptual clarity with logical explanations and solved examples.
4. Makes Class 6 Maths Chapter 2 – Playing with Numbers easier through structured learning.
5. Acts as an excellent reference tool for both guided and self-learning.
6. Reduces exam stress by offering ready-to-practice material in a downloadable format.
7. Helps bridge the gap between theory and application through detailed solutions.