NCERT Solutions Class 6th Maths Chapter – 3 Playing With Numbers Exercise – 3.1

Question 1. Write all the factors of the following numbers:

(a) 24
(b) 15
(c) 21
(d) 27
(e) 12
(f) 20
(g) 18
(h) 23
(i) 36

Solution – 

(a) Factors of 24 are-
ⅰ) 1 × 24 = 24
ⅱ) 2 × 12 = 24
ⅲ) 3 × 8 = 24
ⅳ) 4 × 6 = 24
ⅴ) 6 × 4 = 24
ⅴⅰ) 8 × 3 = 24
ⅵⅰ) 12 × 2 = 24
ⅶⅰ) 24 × 1 = 24

Hence, all the factors of 24 are: 1, 2, 3, 4, 6, 8, 12 and 24.

(b) Factors of 15 are-
ⅰ) 1 × 15 = 15
ⅱ) 3 × 5 = 15
ⅲ) 5 × 3 = 15
ⅳ) 15 × 1 = 15

Hence, all the factors of 15 are: 1, 3, 5 and 15.

(c) Factors of 21 are-
ⅰ) 1 × 21 = 21
ⅱ) 3 × 7 = 21
ⅲ) 7 × 3 = 21
ⅳ) 21 × 1 = 21

Hence, all the factors of 21 are: 1, 3, 7 and 21.

(d) Factors of 27 are-
ⅰ) 1 × 27 = 27
ⅱ) 3 × 9 = 27
ⅲ) 9 × 3 = 27
ⅳ) 27 × 1 = 27

Hence, all the factors of 27 are: 1, 3, 9 and 27.

(e) Factors of 12 are-
ⅰ) 1 × 12 = 12
ⅱ) 2 × 6 = 12
ⅲ) 3 × 4 = 12
ⅳ) 4 × 3 = 12
ⅴ) 6 × 2 = 12
ⅵ) 12 × 1 = 12

Hence, all the factors of 12 are: 1, 2, 3, 4, 6 and 12.

(f) Factors of 20 are-
ⅰ) 1 × 20 = 20
ⅱ) 2 × 10 = 20
ⅲ) 4 × 5 = 20
ⅳ) 5 × 4 = 20
ⅴ) 10 × 2 = 20
ⅵ) 20 × 1 = 20

Hence, all the factors of 20 are: 1, 2, 4, 5, 10 and 20.

(g) Factors of 18 are-
ⅰ) 1 × 18 = 18
ⅱ) 2 × 9 = 18
ⅲ) 3 × 6 = 18
ⅳ) 6 × 3 = 18 
ⅴ) 9 × 2 = 18
ⅵ) 18 × 1 = 18

Hence, all the factors of 18 are: 1, 2, 3, 6, 9 and 18.

(h) Factors of 23 are-
ⅰ) 1 × 23 = 23 
ⅱ) 23 × 1 = 23 

Hence, all the factors of prime number 23 are 1 and 23.

(i) Factors of 36 are-
ⅰ) 1 × 36 = 36 
ⅱ) 2 × 18 = 36 
ⅱ) 3 × 12 = 36 
ⅲ) 4 × 9 = 36  
ⅳ) 6 × 6 = 36
ⅴ) 9 × 4 = 36
ⅵ) 12 × 3 = 36
ⅶ) 18 × 2 = 36 
ⅷ) 36 × 1 = 36

Hence, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.

Question 2. Write first five multiples of-

(a) 5
(b) 8
(c) 9

Solution – 

(a) First five multiples of 5 are-
ⅰ) 5 × 1 = 5
ⅱ) 5 × 2 = 10
ⅲ) 5 × 3 = 15
ⅳ) 5 × 4 = 20
ⅴ) 5 × 5 = 25

Hence, the required multiples of 5 are: 5, 10, 15, 20 and 25.

(b) First five multiples of 8 are-
ⅰ) 8 × 1 = 8
ⅱ) 8 × 2 = 16
ⅲ) 8 × 3 = 24
ⅳ) 8 × 4 = 32
ⅴ) 8 × 5 = 40

Hence, the required multiples of 8 are: 8, 16, 24, 32 and 40.

(c) First five multiples of 9 are-
ⅰ) 9 × 1 = 9
ⅱ) 9 × 2 = 18
ⅲ) 9 × 3 = 27
ⅳ) 9 × 4 = 36
ⅴ) 9 × 5 = 45

Hence, the required multiples of 9 are: 9,18, 27, 36 and 45.

Solution – 

Question 4. Find all the multiples of 9 upto 100.

Solution – 

9 x 1 = 9
9 x 2 = 18
9 x 3 = 27
9 x 4 = 36
9 x 5 = 45
9 x 6 = 54
9 x 7 = 63
9 x 8 = 72
9 x 9 = 81
9 x 10 = 90
9 x 11 = 99
Hence, all the multiples of 9 upto 100 are-
9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.