NCERT Solutions For Class 6 Maths Chapter 3 Playing with Numbers Exercise 3.1 has problems based on factors and multiples. The students can refer to the Solutions prepared by a set of experts at TEXTUAL SOLUTION to practise the NCERT problems given in the textbook effectively. These NCERT Solutions are prepared after conducting vast research on each topic by our experienced faculty.
NCERT Solution for Class 6 Chapter 3 Math : Playing With Numbers Exercise 3.1
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) We have :
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
24 = 6 × 4
Stop here, because 4 and 6 have occured earlier.
Thus, all the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.
(b) We have :
15 = 1 × 15
15 = 3 × 5
15 = 5 × 3
Stop here, because 3 and 5 have occured earlier.
Thus, all the factors of 15 are 1, 3, 5 and 15.
(c) We have :
21 = 1 × 21
21 = 3 × 7
21 = 7 × 3
Stop here, because 3 and 7 have occured earlier.
Thus, all the factors of 21 are 1, 3, 7 and 21.
(d) We have :
27 = 1 × 27
27 = 3 × 9
27 = 9 × 3
Stop here, because 3 and 9 have occured earlier.
Thus, all the factors of 27 are 1, 3, 9 and 27.
(e) We have :
12 = 1 × 12
12 = 2 × 6
12 = 3 × 4
12 = 4 × 3
Stop here, because 3 and 4 have occured earlier.
Thus, all the factors of 12 are 1, 2, 3, 4, 6 and 12.
(f) We have :
20 = 1 × 20
20 = 2 × 10
20 = 4 × 5
20 = 5 × 4
Stop here, because 4 and 5 occured earlier.
Thus, all the factors of 20 are 1, 2, 4, 5, 10 and 20.
(g) We have :
18 = 1 × 18
18 = 2 × 9
18 = 3 × 6
18 = 6 × 3
Stop here, because 3 and 6 have occured earlier.
Thus, all the factors of 18 are 1, 2, 3, 6, 9 and 18.
(h)We have :
23 = 1 × 23
23 = 23 × 1
Stop here, because 1 and 23 have occured earlier.
Thus, all the factors of 23 are 1 and 23.
(i) We have :
36 = 1 × 36
36 = 2 × 18
36 = 3 × 12
36 = 4 × 9
36 = 6 × 6
Stop here, because both the factors (6) are same.
Thus, all the factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36.
2. Write first five multiples of :
(a) 5 (b) 8 (c) 9
Solutions :
(a) In order to obtain first five multiples of 5, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
5 × 1 = 5
5 × 2 = 10
5 × 3 = 15
5 × 4 = 20
5 × 5 = 25
Hence, the first five multiples of 5 are 5, 10, 15, 20 and 25 respectively.
(b) In order to obtain first five multiples of 8, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
8 × 1 = 8
8 × 2 = 16
8 × 3 = 24
8 × 4 = 32
8 × 5 = 40
Hence, the first five multiples of 8 are 8, 16, 24, 32 and 40 respectively.
(c) In order to obtain first five multiples of 9, we multiply it by 1, 2, 3, 4 and 5 respectively.
We have,
9 × 1 = 9
9 × 2 = 18
9 × 3 = 27
9 × 4 = 36
9 × 5 = 45
Hence, the first five multiples of 9 are 9, 18, 27, 36 and 45 respectively.
3. Match the items in column 1 with the items in column 2.
| Column 1 | Column 2 |
|---|---|
| (i) 35 | (a) Multiple of 8 |
| (ii) 15 | (b) Multiple of 7 |
| (iii) 16 | (c) Multiple of 70 |
| (iv) 20 | (d) Factor of 30 |
| (v) 25 | (e) Factor of 50 |
| (f) Factor of 20 |
| Column 1 | Column 2 |
|---|---|
| (i) 35 | (b) Multiple of 7 |
| (ii) 15 | (d) Factor of 30 |
| (iii) 16 | (a) Multiple of 8 |
| (iv) 20 | (f) Factor of 20 |
| (v) 25 | (e) Factor of 50 |
4. Find all the multiples of 9 upto 100.
Solution :
All the multiples of 9 upto 100 are :
9 × 1, 9 × 2, 9 × 3, 9 × 4, 9 × 5, 9 × 6, 9 × 7, 9 × 8, 9 × 9, 9 × 10 and 9 × 11.
i.e., 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.