Chapter - 3 Playing with Numbers

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Q
Playing With Numbers

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) 24 can be written as :

24 = 1 x 24,    24 = 2 x 12,     24 = 3 x 8,     24 = 4 x 6,   24 = 6 x 4

Since, 6 and 4 have occurred earlier, we can stop here.

So, the factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24.

 

(b) 15 can be written as:

15 = 1 x 5,    15 = 3 x 5,     15 = 5 x 3

Since, 5 and 3 have occurred earlier, we can stop here.

So, the factors of 15 are1, 3, 5 and 15.

 

(c) 21 can be written as:

21 = 1 x 21,    21 = 3 x 7,     21 = 7 x 3

Since 7 and 3 have occurred earlier, we can stop here.

So, the factors of 21 are 1, 3, 7 and 21.

 

(d) 27 can be written as:

27 = 1 x 27,   27 = 3 x 9,    27 = 9 x 3

Since, 9 and 3 have occurred earlier, we can stop here.

So, the factors of 27 are 1, 3, 9 and 27.

 

(e) 12 can be written as:

12 = 1 x 12,   12 = 2 x 6,   12 = 3 x 4,   12 = 4 x 3

Since, 3 and 4 have occurred earlier, we can stop here.

So, the factors of 12 are 1, 2, 3, 4, 6 and 12.

 

(f) 20 can be written as:

20 = 1 x 20,    20 = 2 x 10,     20 = 4 x 5,    20 = 5 x 4

Since, 5 and 4 have occurred earlier, we can stop here.

So, the factors of 20 are 1, 2, 4, 5, 10 and 20.

 

(g) 18 can be written as:

18 = 1 x 18,   18 = 2 x 9,    18 = 3 x 6,    18 = 6 x 3

Since, 6 and 3 have occurred earlier, we can stop here.

So, the factors of 18 are 1, 2, 3, 6, 9 and 18.

 

(h) 23 can be written as:

23 = 1 x 23,     23 = 23 x 1

Since, 23 and 1 have occurred earlier, we can stop here.

So, the factors of 23 are 1 and 23.

 

(i) 36 can be written as:

36 = 1 x 36,    36 = 2 x 18,    36 = 3 x 12,    36 = 4 x 9,   36 = 6 x 6

We can stop here as both the factors are same (6).

So, the factor 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

 

Solution: 

(a) The multiples are:

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

 

(b) The multiples are:

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

 

c) The multiples are:

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.

 

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

 

Solution: 

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: Multiples of 9 upto 100 are:

9 × 1 = 9,   9 × 2 = 18,    9 × 3 = 27,   9 × 4 = 36,    9 × 5 = 45,   9 × 6 = 54,    9 × 7 = 63

9 × 8 = 72,    9 × 9 = 81,   9 × 10 = 90,    9 × 11 = 99

Thererfore, all the multiples of 9 upto 100 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90 and 99.

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