## SOLUTIONS FOR CBSE CLASS 6 NCERT MATHS CHAPTER 3 | EX – 3.4 & EX – 3.5

Find here, the step-by-step and precisely written Solutions for CBSE Class 6 NCERT Maths Chapter 3 Playing with Numbers | Exercise 3.4 and Exercise 3.5.

Before moving to Solutions For CBSE Class 6 NCERT Maths Chapter 3, it is advised to go through the following topics:
• Playing With Numbers

## Solutions For CBSE Class 6 NCERT Maths Chapter 3 | Exercise 3.4

Class 6 Exercise 3.4 deals with Common Factors and Common Multiples of two or more numbers.

Co-primes : If two numbers have no common factors other than 1 between them then they are called Co-primes.

### Common Factors | Common Multiples

QUESTION 1

Find the common factors of

(i) 20 and 28 (ii) 15 and 25
(iii) 35 and 50 (iv) 56 and 120

ANSWER

 (i) 1 × 20 = 20 2 × 10 = 20 4 × 5 = 20 Factors of 20 = 1, 2, 4, 5, 10 and 20 1 × 28 = 28 2 × 14 = 28 4 × 7 = 28 Factors of 28 = 1, 2, 4, 7, 14 and 28

Common factors of 20 and 28 = 1, 2 and 4

 (ii) 1 × 15 = 15 3 × 5 = 15 Factors of 15 = 1, 3, 5 and 15 1 × 25 = 25 5 × 5 = 25 Factors of 25 = 1, 5 and 25

Common factors of 15 and 25 = 1 and 5

 (iii) 1 × 35 = 35 5 × 7 = 35 Factors of 35 = 1, 5, 7 and 20 1 × 50 = 50 2 × 25 = 50 5 × 10 = 50 Factors of 50 = 1, 2, 5, 10, 25 and 50

Common factors of 35 and 50 = 1 and 5

 (iv) 1 × 56 = 56 2 × 28 = 56 4 × 14 = 56 7 × 8 = 56 Factors of 56 = 1, 2, 4, 7, 8, 14, 28 and 56 1 × 120 = 120 2 × 60 = 120 3 × 40 = 120 4 × 30 = 120 5 × 24 = 120 6 × 20 = 120 8 × 15 = 120 10 × 12 = 120 Factors of 120 = 1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60 and 120

Common factors of 56 and 120 = 1, 2, 4 and 8.

QUESTION 2

Find the common factors of:
(i) 4, 8 and 12 (ii) 5, 15 and 25

ANSWER

 (i) 1 × 4 = 4 2 × 2 = 4 Factors of 4 = 1, 2 and 4 1 × 8 = 8 2 × 4 = 8 Factors of 8 = 1, 2, 4 and 8 1 × 12 = 12 2 × 6 = 12 3 × 4 = 12 Factors of 12 = 1, 2, 3, 4, 6 and 12

Common factors of 4, 8 and 12 = 1, 2 and 4

 (ii) 1 × 5 = 5 Factors of 5 = 1 and 5 1 × 15 = 15 3 × 5 = 15 Factors of 8 = 1, 3, 5 and 15 1 × 25 = 25 5 × 5 = 25 Factors of 25 = 1, 5 and 25

Common factors of 5, 15 and 25 = 1 and 5.

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.4 ]

QUESTION 3

Find the first three common multiples of

(i) 6 and 8 (ii) 12 and 18

ANSWER

 (i) 6 × 1 = 6, 6 × 2 = 12, 6  × 3 = 18, 6 × 4 = 24, 6 × 5 = 30, 6 × 6 = 36, 6 × 7 = 42, 6 × 8 = 48, 6 × 9 = 54, 6 × 10 = 60, 6 × 11 = 66, 6 × 12 = 72, … Multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 28, 54, 60, 72, … 8 × 1 = 8, 8 × 2 = 16, 8 × 3 = 24, 8 × 4 = 32, 8 × 5 = 40, 8 × 6 = 48, 8 × 7 = 56, 8 × 8 = 64, 8 × 9 = 72, 8 × 10 = 80, … Multiples of 8 = 8, 16, 24, 32, 40, 48, 56, 64, 72, …

Thus, First three Common multiples of 6 and 8 = 24, 48 and 72.

 (ii) 12 × 1 = 12, 12 × 2 = 24, 12 × 3 = 36, 12 × 4 = 48, 12 × 5 = 60, 12 × 6 = 72, 12 × 7 = 84, 12 × 8 = 96, 12 × 9 = 108, 12 × 10 = 120, … Multiples of 12 = 12, 24, 36, 48, 60, 72, 84, 96, 108, 120, … 18 × 1 = 18, 18 × 2 = 36, 18 × 3 = 54, 18 × 4 = 72, 18 × 5 = 90, 18 × 6 = 108, … Multiple of 18 = 18, 36, 54, 72, 90, 108, …

QUESTION 4

Write all the numbers less than 100 which are common multiples of 3 and 4.

ANSWER

Multiples of 3 less than 100 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60, 63, 66, 69, 72, 75, 78, 81, 84, 87, 90, 93, 96, 99

Multiple of 4 less than 100 = 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96

Therefore, Common multiples of 3 and 4 less than 100 = 12, 24, 36, 48, 60, 72, 84 and 96.

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.4 ]

QUESTION 5

Which of the following numbers are co-primes:

(i) 18 and 35 (ii) 15 and 37

(iii) 30 and 415 (iv) 17 and 68

(v) 216 and 215 (vi) 81 and 16

ANSWER

 (i) 1 × 18 = 18 2 × 9 = 18 3 × 6 = 18 Factors of 18 = 1, 2, 3, 6, 9 and 18 1 × 35 = 35 5 × 7 = 35 Factors of 35 = 1, 5, 7 and 35

Common factors of 18 and 35 = 1

Since, 18 and 35 have only one common factor, i.e., 1, therefore, they are co-primes.

 (ii) 1 × 15 = 15 3 × 5 = 15 Factors of 15 = 1, 3, 5, and 15 1 × 37 = 37 Factors of 37 = 1 and 37

Common factor of 15 and 37 = 1

Since, 15 and 37 have only one common factor, i.e., 1, therefore, they are co-primes

 (iii) 1 × 30 = 30 2 × 15 = 30 3 × 10 = 30 5 × 6 = 30 Factors of 30 = 1, 2, 3, 5, 6, 10, 15 and 30 1 × 415 = 415 5 × 83 = 415 Factors of 415 = 1, 5, 83 and 415

Common factor of 30 and 415 = 1 and 5

Since, 30 and 415 have more than one common factor, therefore, they are not co-primes.

 (iv) 1 × 17 = 17 Factors of 17 = 1 and 17 1 × 68 = 68 2 × 34 = 68 4 × 17 = 68 Factors of 68 = 1, 2, 4, 17, 34 and 68

Common factor = 1 and 17

Since, 17 and 68 have more than one common factor, therefore, they are not co-primes.

 (v) 1 × 216 = 216 2 × 108 = 216 4 × 54 = 216 6 × 36 = 216 8 × 27 = 216 9 × 24 = 216 12 × 18 = 216 Factors of 216 = 1, 2, 4, 6, 8, 9, 12, 18, 24, 27, 36, 54, 108 and 216 1 × 215 = 215 5 × 43 = 215 Factors of 215= 1, 5, 43 and 215

Common factor of 215 and 216 = 1

Since, both have only one common factor, i.e., 1, therefore, they are co-primes.

 (vi) 1 × 81 = 81 3 × 27 = 81 9 × 9 = 81 Factors of 81 = 1, 3, 9, 27 and 81 1 × 16 = 16 2 × 8 = 16 4 × 4 = 16 Factors of 16 = 1, 2, 4, 8 and 16

Common factor of 81 and 16 = 1

Since, 81 and 16 have only one common factor, i.e., 1, therefore, they are co-prime.

❔ What are co-primes?
❔ What are twin primes?

QUESTION 6

A number is divisible by both 5 and 12. By which other number will that number be always divisible?

ANSWER

5 × 12 = 60.

The number will be divisible by 60.

ℹ A number divisible by two co-prime numbers will be divisible by their product also.

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.4 ]

QUESTION 7

A number is divisible by 12. By what other numbers will that number be divisible?

ANSWER

A number is divisible by 12 then it is divisible by each of the factors of 12.

Factors of 12 = 1, 2, 3, 4, 6 and 12.

Thus, the number will also be divisible by 1, 2, 3, 4 and 6.

ℹ A number divisible by another number will always be divisible by each of the factors of that number.

## Solutions For CBSE Class 6 NCERT Maths Chapter 3 | Exercise 3.5

Class 6 Exercise 3.5 deals with Prime Factorisation Method and it’s application in finding Divisibility of Numbers.

### EXERCISE 3.5 Factorisation

QUESTION 1

Select the true statements:

(i) If a number is divisible by 3, it must be divisible by 9.
– False

(ii) If a number is divisible by 9, it must be divisible by 3.
-True

(iii) If a number is divisible by 18, it must be divisible by both 3 and 6.
– True

(iv) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
– True

(v) If two numbers are co-primes, at least one of them must be prime.
– False

(vi) All numbers which are divisible by 4 must also by divisible by 8.
– False

(vii) All numbers which are divisible by 8 must also by divisible by 4.
– True

(viii) If a number is exactly divides two numbers separately, it must exactly divide their sum.
– True

(ix) If a number is exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
– False

QUESTION 2

Here are two different factor trees for 60. Write the missing number.

 (i)  (ii)  ANSWER

 (i)  (ii)  [ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 3

Which factors are not included in the prime factorization of a composite number?

ANSWER

1

QUESTION 4

Write the greatest 4-digit number and express it in terms of its prime factors.

ANSWER

The greatest four digit number = 9999 9999 = 3 × 3 × 11 × 101

❔ What is prime factorisation?
❔ How to do prime factorisation of a number by division method?

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 5

Write the smallest 5-digit number and express it in terms of its prime factors.

ANSWER

The smallest five digit number = 10000. 10000 = 2 × 2 × 2 × 2 × 5 × 5 × 5 × 5

QUESTION 6

Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any, between, two consecutive prime factors.

ANSWER 1729 = 7 × 13 × 19

When arranged in ascending order, the difference of two consecutive prime factors is 6.

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 7

The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

ANSWER

(i) 3 × 4 × 5 = 60, divisible by 6

(ii) 8 × 9 × 10 = 720, divisible by 6.

(iii) 13 × 14 × 15 = 2730 (= 455 × 6)

ℹ Even number is divisible by 2 and number which is a multiple of 3 is divisible by 3.

Among any three consecutive numbers, one number must always be even and one number must always be a multiple of 3. (Since, every third natural number is a multiple of 3)
Then,
Product of three consecutive numbers = (where k, m and n are whole numbers)
= => Product of three consecutive numbers is a multiple of 6.

Hence, the product three consecutive numbers is always divisible by 6.

❔ Why product of three consecutive numbers is always divisible by 6?

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 8

The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

ANSWER

3 + 5 = 8, divisible by 4.

7 + 9 = 16, divisible by 4.

13 + 15 = 28, divisible by 4.

19 + 21 = 40, divisible by 4.

ℹ An Even number is divisible by 2
=> 2n is an even number, where n is a whole number.
=> 2n + 1 is an odd number.

Sum of two consecutive odd numbers = (2n + 1) + (2n + 3)
= 4n + 4
= 4(n + 1)
=> Sum of two consecutive odd numbers is divisible by 4.

❔ Why sum of two consecutive odd numbers is always divisible by 4?

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 9

In which of the following expressions, prime factorization has been done:

(i) 24 = 2 × 3 × 4

(ii) 56 = 7 × 2 × 2 × 2

(iii) 70 = 2 × 5 × 7

(iv) 54 = 2 × 3 × 9

ANSWER

In prime factorisation all factors are prime numbers. Hence, in expressions (ii) and (iii), prime factorization has been done.

QUESTION 10

Determine if 25110 is divisible by 45.

[Hint: 5 and 9 are co-prime numbe Test the divisibility of the number by 5 and 9]

ANSWER

25110 is divisible by 5 as ‘0’ is at its unit place.

2 + 5 + 1 + 1 + 0 = 9
25110 is divisible by 9 as sum of digits is divisible by 9.

Factors of 5 = 1 and 5,
Factors of 9 = 1, 3 and 9
Common factors of 5 and 9 = 1
=> 5 and 9 are coprimes.

If a number is divisible by co-primes then it is divisible by their product also.
=> 25110 is divisible by 5×9 = 45.

❔ Why a number divisible by a pair of co-primes is also divisible by their product?

[ Solutions For CBSE Class 6 NCERT Maths Chapter 3 Exercises 3.5 ]

QUESTION 11

18 is divisible by both 2 and 3. It is also divisible by 2 × 3 = 6. Similarly, a number is divisible by 4 and 6. Can we say that the number must be divisible by 4 × 6 = 24? If not, give an example to justify your answer.

ANSWER

If a number is divisible by a pair of numbers and their products also then the pair of numbers must be coprimes.

18 is divisible by 2 and 3. It is also divisible by (2 × 3 =) 6 because 2 and 3 are coprimes.

A number divisible by 4 and 6 may not be divisible by (4 × 6 =) 24 because 4 and 6 are not coprimes.

e.g., 36 is divisible by both 4 and 6 but 36 is not divisible by 24.

QUESTION 12

I am the smallest number, having four different prime factors. Can you find me?

ANSWER

Four different prime numbers in order starting with the smallest
= 2, 3, 5 and 7.
Therefore, the smallest number having four different prime factors = 2 × 3 × 5 × 7
= 210.

## NCERT MATHS SOLUTION | CLASS 6 | CHAPTER 3 | PLAYING WITH NUMBERS

Exercise 3.1 | Exercise 3.2 | Exercise 3.3

Exercise 3.6 | Exercise 3.7