Class 6 Maths Chapter 3 Playing with Numbers

NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Factors and Multiples

Q Find the possible factors of 45, 30 and 36.

Answer: The possible factor of

45=1,3,5,9,15,45
30=1,2,3,5,6,10,15,30
36=1,2,3,4,6,9,12,18,36

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.1

Q1 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

Answer: (a) 24 = 1 $\times$ 24

= 2 $\times$ 12

= 3 $\times$ 8

= 4 $\times$ 6

Hence Factor of 24 = 1, 2, 3, 4, 6, 8, 12 and 24 itself.

(b) 15 = 1 $\times$ 15

= 3 $\times$ 5

= 5 $\times$ 3

Hence factor of 15 = 1, 3, 5 and 15.

= 3 $\times$ 7

= 7 $\times$ 3

Hence factor of 21 = 1, 3, 7 and 21.

(d) 27 = 1 $\times$ 27

= 3 $\times$ 9

= 9 $\times$ 3

Hence Factor of 27 are 1, 3, 9 and 27.

(e) 12=1 $\times$ 12

=2 $\times$ 6

=3 $\times$ 4

=4 $\times$ 3

Hence Factors of 12 are 1,2,3,4,6, and 12
(f) 20 = 1 $\times$ 20

= 2 $\times$ 10

= 4 $\times$ 5

= 5 $\times$ 4

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

(g) 18 = 1 $\times$ 18

= 2 $\times$ 9

= 3 $\times$ 6

Hence Factors of 18 are 1, 2, 3, 6, 9 and 18.

(h) 23 = 1 $\times$ 23

= 23 $\times$ 1

Hence factors of 23 are 1 and 23.

(i) 36 = 1 $\times$ 36

= 2 $\times$ 18

= 3 $\times$ 12

= 4 $\times$ 9

= 6 $\times$ 6

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

Q2 Write first five multiples of :
(a) 5 (b) 8 (c) 9

Answer: First Five multiple of

(a) 5

5 $\times$ 1 = 5

5 $\times$ 2 = 10

5 $\times$ 3 = 15

5 $\times$ 4 = 20

5 $\times$ 5 = 25

(b) 8

8 $\times$ 1 = 8

8 $\times$ 2 = 16

8 $\times$ 3 = 24

8 $\times$ 4 = 32

8 $\times$ 5 = 40

9 $\times$ 1 = 9

9 $\times$ 2 = 18

9 $\times$ 3 = 27

9 $\times$ 4 = 36

9 $\times$ 5 = 45

Q3 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

Answer: Column 1 Column 2

(i) 35 (b) Multiple of 7

(ii) 15 (d) Factor of 30

(iii) 16 (a) Multiple of 8

(iv) 20 (d) Factor of 20

(v) 25 (e) Factor of 50

Q4 Find all the multiples of 9 up to 100.

Answer: Multiples of 9 up to 100 are:

9 $\times$ 1 = 9

9 $\times$ 2 = 18

9 $\times$ 3 = 27

9 $\times$ 4 = 36

9 $\times$ 5 = 45

9 $\times$ 6 = 54

9 $\times$ 7 = 63

9 $\times$ 8 = 72

9 $\times$ 9 = 81

9 $\times$ 10 = 90

9 $\times$ 11 = 99

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime and Composite Numbers

Q1 Observe that 2 × 3 + 1 = 7 is a prime number. Here, 1 has been added to a multiple of 2 to get a prime number. Can you find some more numbers of this type?

Answer: some more numbers of this type are :

2 $\times$ 2+1=5
2 $\times$ 5+1=11
2 $\times$ 6+1=13

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.2

Q1 What is the sum of any two (a) Odd numbers? (b) Even numbers?

Answer: (a) the sum of any two Odd numbers is always even. for e.g. 3 + 5 = 8 and 9 + 7 = 16

(b) the sum of any two Even numbers is always even. for e.g. 2 + 4 = 6 and 8 + 4 = 12

Q2 State whether the following statements are True or False:
(a) The sum of three odd numbers is even.
(b) The sum of two odd numbers and one even number is even.
(c) The product of three odd numbers is odd.
(d) If an even number is divided by 2, the quotient is always odd.
(e) All prime numbers are odd.
(f) Prime numbers do not have any factors.
(g) The sum of two prime numbers is always even.
(h) 2 is the only even prime number.
(i) All even numbers are composite numbers.
(j) The product of two even numbers is always even.

Answer: (a) False, Because the sum of two odd numbers is even and the sum of even number and an odd number is odd, so the sum of three odd numbers is odd.

for example :

3 + 5 + 7 = 15, i.e., odd

(b) True, as the sum of two odd number is even a sum of two even number is even.

for example :

3 + 5 + 6 = 14, i.e., even

(c) True because the product of two odd numbers is odd and product of any number(odd or even) with an even number is even.

For example :

3 x 5 x 7 = 105, i.e., odd

(d) False, because it is possible to have a quotient even when divided by 2

for example :

4÷2=24÷2=2, i.e., even

(e) False as 2 is a prime number and it is also even

(f) False as 1 and the number itself are factors of the number

(g) False for example

2 + 3 = 5 , i.e., odd

(h) True

(i) False as 2 is a prime number

(j) True .

Q3 The numbers 13 and 31 are prime numbers. Both these numbers have the same digits 1 and 3. Find such pairs of prime numbers up to 100.

Answer: Prime Number having the same digit are:

17,71

37,73 and

79,97

Q4 Write down separately the prime and composite numbers less than 20.

Answer: Prime numbers less than 20 are : 2, 3, 5, 7, 11, 13, 17, 19

Composite numbers less than 20 are : 4, 6, 8, 9, 10, 12, 14, 15, 16, 18

Q5 What is the greatest prime number between 1 and 10?

Answer: Prime numbers between 1 and 10 are 2, 3, 5, and 7. Among these numbers, 7 is the greatest.

Q6 Express the following as the sum of two odd primes.
(a) 44 (b) 36 (c) 24 (d) 18

Answer: (a) 44 = 37 +7

(b) 36 = 31 +5

(d) 18 = 11 +7

Q7 Give three pairs of prime numbers whose difference is 2.
[Remark: Two prime numbers whose difference is 2 are called twin primes].

Answer: three pairs of prime numbers whose difference is 2 are :

3,5

41,43 and

71,73

Q8 Which of the following numbers are prime?
(a) 23 (b) 51 (c) 37 (d) 26

Answer: (a) 23,23=1 $\times$ 23,23=23 $\times$ 123,23=1 $\times$ 23,23=23 $\times$ 1 23 has only two factors, 1 and 23. Therefore, it is a prime number.

(b) 51,51=1 $\times$ 51,51=3 $\times$ 1751,51=1 $\times$ 51,51=3 $\times$ 17 51 has four factors, 1, 3, 17, 51. Therefore, it is not a prime number. It is a composite number.

(d) 26 26 has four factors (1, 2, 13, 26). Therefore, it is not a prime number. It is a composite number.

Q9 Write seven consecutive composite numbers less than 100 so that there is no prime number between them.

Answer: Seven consecutive composite numbers less than 100 so that there is no prime number between them are :

Between 89 and 97, both of which are prime numbers, there are 7 composite numbers. They are: 90,91,92,93,94,95,96

Q10 Express each of the following numbers as the sum of three odd primes:
(a) 21 (b) 31 (c) 53 (d) 61

Answer: (a) 21 = 3 + 7 + 11

(b) 31 = 5 + 7 + 19

(d) 61 = 11 + 19 + 31

Q11 Write five pairs of prime numbers less than 20 whose sum is divisible by 5. (Hint : 3+7 = 10)

Answer: Five pairs of prime numbers less than 20 whose sum is divisible by 5 are :

2+3=5

2+13=15

3+17=20

7+13=20

19+11=30

Q12 Fill in the blanks :
(a) A number that has only two factors is called a ______.
(b) A number that has more than two factors is called a ______.
(c) 1 is neither ______ nor ______.
(d) The smallest prime number is ______.
(e) The smallest composite number is _____.
(f) The smallest even number is ______.

Answer: (a) prime number

(b) Composite number

(d) 2

(e) 4

(f) 2

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.3

Q1 Using divisibility tests, determine which of the following numbers are divisible by 2; by 3; by 4; by 5; by 6; by 8; by 9; by 10 ; by 11 (say, yes or no):

Number	Divisible by
Number	2	3	4	5	6	8	9	10	11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	......	......	......	......	......	......	......	......	......
1586	......	......	......	......	......	......	......	......	......
275	......	......	......	......	......	......	......	......	......
6686	......	......	......	......	......	......	......	......	......
639210	......	......	......	......	......	......	......	......	......
429714	......	......	......	......	......	......	......	......	......
2856	......	......	......	......	......	......	......	......	......
3060	......	......	......	......	......	......	......	......	......
406839	......	......	......	......	......	......	......	......	......

Answer:

Number	Divisible by
Number	2	3	4	5	6	8	9	10	11
128	Yes	No	Yes	No	No	Yes	No	No	No
990	Yes	Yes	No	Yes	Yes	No	Yes	Yes	Yes
1586	Yes	No	No	No	No	No	No	No	No
275	No	No	No	Yes	No	No	No	No	Yes
6686	Yes	No	No	No	No	No	No	No	No
639210	Yes	Yes	No	Yes	Yes	No	No	Yes	Yes
429714	Yes	Yes	No	No	Yes	No	Yes	No	No
2856	Yes	Yes	Yes	No	Yes	Yes	No	No	No
3060	Yes	Yes	Yes	Yes	Yes	No	Yes	Yes	No
406839	No	Yes	No	No	No	No	No	No	No

Q2 Using divisibility tests, determine which of the following numbers are divisible by 4; by 8:
(a) 572 (b) 726352 (c) 5500 (d) 6000 (e) 12159
(f) 14560 (g) 21084 (h) 31795072 (i) 1700 (j) 2150

Answer: A number with 3 or more digits is divisible by 4 if the number formed by its last two digits is divisible by 4.

A number with 3 or more digits is divisible by 8 if the number formed by its last three digits is divisible by 8.

a) 572

72 is divisible by 4, hence the number is divisible by 4.

The number is not divisible by 8.

b) 726352

52 is divisible by 4, hence the number is divisible by 4.

352 is divisible by 8, hence the number is divisible by 8.

c) 5500

0 is divisible by 4, hence the number is divisible by 4.

500 is not divisible by 8, hence the number is not divisible by 8.

d) 6000

0 is divisible by 4 and 8. Hence, the number is divisible by 4 and 8.

e) 12159

59 is not divisible by 4. Hence, the number is not divisible by 4.

159 is not divisible by 8, hence the number is not divisible by 8.

f) 14560

60 is divisible by 4, hence the number is divisible by 4.

560 is divisible by 8, hence the number is divisible by 8.

g) 21084

84 is divisible by 4, hence the number is divisible by 4.

84 is not divisible by 8, hence the number is not divisible by 8.

h) 31795072

72 is divisible by 4 and 8. Hence the number is divisible by 4 and 8.

i) 1700

The number is divisible by 4.

700 is not divisible by 8, hence the number is not divisible by 8.

j) 2150

50 is not divisible by 4, hence the number is not divisible by 4.

150 is not divisible by 8, hence the number is not divisible by 8.

Q3 Using divisibility tests, determine which of the following numbers are divisible by 6:
(a) 297144 (b) 1258 (c) 4335 (d) 61233 (e) 901352
(f) 438750 (g) 1790184 (h) 12583 (i) 639210 (j) 17852

Answer: (a) 297144

Since the last digit Of the number is 4, it is divisible by 2. On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3. As the number is divisible by both 2 and 3, it is divisible by 6.

(b) 1258

Since the last digit of the number is 8, it is divisible by 2. On adding all the digits of the number, the sum obtained is 16. Since 16 is not divisible by 3, the given number is also not divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

The last digit of the number is 5, which is not divisible by 2. Therefore, the given number is also not divisible by 2. On adding all the digits of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

(d) 61233

The last digit of the number is 3, which is not divisible by 2. Therefore, the given number is also not divisible by 2. On adding all the digits Of the number, the sum obtained is 15. Since 15 is divisible by 3, the given number is also divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

(e) 901352

Since the last digit of the number is 2, it is divisible by 2. On adding all the digits of the number, the sum obtained is 20. Since 20 is not divisible by 3, the given number is also not divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

(f) 438750

Since the last digit of the number is O, it is divisible by 2. On adding all the digits of the number, the sum obtained is 27. Since 27 is divisible by 3, the given number is also divisible by 3. As the number is divisible by both 2 and 3, it is divisible by 6.

(g) 1790184

Since the last digit of the number is 4, it is divisible by 2. On adding all the digits of the number, the sum obtained is 30. Since 30 is divisible by 3, the given number is also divisible by 3. As the number is divisible by both 2 and 3, it is divisible by 6.

(h) 12583

Since the last digit of the number is 3, it is not divisible by 2. On adding all the digits of the number, the sum obtained is 19. Since 19 is not divisible by 3, the given number is also not divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

(i) 639210

Since the last digit of the number is O, it is divisible by 2. On adding all the digits of the number, the sum obtained is 21. Since 21 is divisible by 3, the given number is also divisible by 3. As the number is divisible by both 2 and 3, it is divisible by 6.

(j) 17852

Since the last digit of the number is 2, it is divisible by 2. On adding all the digits of the number, the sum obtained is 23. Since 23 is not divisible by 3, the given number is also not divisible by 3. As the number is not divisible by both 2 and 3, it is not divisible by 6.

Q4 Using divisibility tests, determine which of the following numbers are divisible by 11:
(a) 5445 (b) 10824 (c) 7138965 (d) 70169308 (e) 10000001
(f) 901153

Answer: (a) 5445

Sum of the digits at odd places = 5 + 4 = 9 Sum of the digits at even places = 4 + 5 = 9 Difference = 9 - 9 = 0 As the difference between the sum of the digits at odd places and the sum of the digits at even places is O, therefore, 5445 is divisible by 11.

(b) 10824

Sum of the digits at odd places = 4 + 8 + 1 = 13 Sum of the digits at even places = 2 + 0 = 2 Difference = 13 - 2 = 11 The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 10824 is divisible by 11.

Sum of the digits at odd places = 5 + 9 + 3 + 7 = 24 Sum of the digits at even places =6 + 8 + 1 = 15 Difference = 24 - 15 = 9 The difference between the sum of the digits at odd places and the sum of digits at even places is 9, which is not divisible by 11. Therefore, 7138965 is not divisible by 11.

(d) 70169308

Sum of the digits at odd places = 8 + 3 + 6 + 0 Sum of the digits at even places = 0 + 9 + 1 + 7 = 17 Difference = 17 - 17 = 0 As the difference between the sum of the digits at odd places and the sum of the digits at even places is O, therefore, 70169308 is divisible by 11.

(e) 10000001

Sum Of the digits at Odd places = 1 Sum of the digits at even places 1 Difference = 1 - 1 = 0 As the difference between the sum of the digits at odd places and the sum of the digits at even places is O, therefore, 10000001 is divisible by 11.

(f) 901153

Sum of the digits at odd places = 3 + 1 + 0 = 4 Sum Of the digits at even places = 5 + 1 + 9 = 15 Difference = 15 - 4 = 11 The difference between the sum of the digits at odd places and the sum of the digits at even places is 11, which is divisible by 11. Therefore, 901153 is divisible by 11.

Q5 Write the smallest digit and the greatest digit in the blank space of each of the following numbers so that the number formed is divisible by 3 : (a) __ 6724 (b) 4765 __ 2

Answer: (a) _6724

Sum of the remaining digits = 19

To make the number divisible by 3, the sum of its digits should be divisible by 3. The smallest multiple of 3 which comes after 19 is 21.

Therefore,

smallest number = 21 - 19 - 2

Now, 2 + 3 + 3 = 8 If we put 8, then the sum of the digits will be 27 and as 27 is divisible by 3, the number will also be divisible by 3.

Therefore, the largest number is 8.

(b) 4765_2

Sum of the remaining digits = 24

To make the number divisible by 3, the sum of its digits should be divisible by 3. As 24 is already divisible by 3, the smallest number that can be placed here is 0.

Now, 0 + 3 = 3 3 + 3 = 6 3 + 3 + 3 = 9

How ever, 3 + 3 + 3 + 3 = 12

If we put 9, then the sum of the digits will be 33 and as 33 is divisible by 3, the number will also be divisible by 3.

Therefore, the largest number is 9.

Q6 Write a digit in the blank space of each of the following numbers so that the number formed is divisible by 11 : (a) 92 __ 389 (b) 8 __ 9484.

Answer: (a) 92 __ 389

Sum of odd digits = 9 + (blank space) + 8 = 17 + blank space

Sum of even digits = 2 + 3 + 9 = 14

As we know,

The number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places is divisible by 11.

If we make the sum of odd digits = 25

then we will have difference = 25 - 14 = 11

which is divisible by 11.

To make the sum of odd digits = 25,

the number at black space would be 8.

(b) 8 __ 9484

Sum of odd digits = 8 + 9 + 8 = 25

Sum of even digits = blank space + 4 + 4 = blank space + 8

The number is divisible by 11 if the difference between the sum of the digits at odd places and the sum of the digits at even places is divisible by 11.

If we make the sum of even digits = 14 then we will have difference = 25 - 14 = 11 which is divisible by 11.

To make the sum of even digits = 14,

the number at black space would be 6.

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Common Factors and Common Multiples

Q Find the common factors of:
(a) 8, 20 (b) 9, 15

Answer: The common factors of the following are:
(a) 8, 20

$8 = 2\times2\times2\times2$

$20 = 2\times2\times5$

Hence, the common factors are $1, 2\ and\ (2\times2 = )4$

(b) 9, 15

$9 = 3\times3$

$15= 3\times5$

Hence, the common factors are $1\ and\ 3$ .

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.4

Q1 Find the common factors of :
(a) 20 and 28 (b) 15 and 25 (c) 35 and 50 (d) 56 and 120

Answer: (a) 20 and 28

Factors of 20=1,2,4,5,10,20

Factors of 28=1,2,4,7,14,28

Common factors =1,2,4

(b) 15 and 25

Factors of 15=1,3,5,15

Factors of 25=1,5,25

Common factors =1,5

Factors of 35=1,5,7,35

Factors of 50=1,2,5,10,25,50

Common factors =1,5

(d) 56 and 120

Factors of 56=1,2,4,7,8,14,28,56

Factors of 120=1,2,3,4,5,6,8,10,12,15,20,24,30,40,60,120

Common factors =1,2,4,8

Q2 Find the common factors of :
(a) 4, 8 and 12 (b) 5, 15 and 25

Answer: (a) 4,8,12

Factors of 4=1,2,4

Factors of 8=1,2,4,8

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

Common factors =1,2,4

(b) 5,15, and 25

Factors of 5=1,5

Factors of 15=1,3,5,15

Factors of 25=1,5,25

Common factors =1,5

Q3 Find the first three common multiples of :
(a) 6 and 8 (b) 12 and 18

Answer: (a) 6 and 8

Multiple of 6=6,12,18,24,30…

Multiple of 8=8,16,24,32……

common multiples =24,48,72

(b) 12 and 18

Multiples of 12=12,24,36,78

Multiples of 18=18,36,54,72 3

common multiples = 36,72,108

Q4 Write all the numbers less than 100 which are common multiples of 3 and 4.

Answer: Multiples of 3 = 3,6,9,12,15…

Multiples of 4 = 4,8,12,16,20…

Common multiples = 12,24,36,48,60,72,84,96

Q5 Which of the following numbers are co-prime?
(a) 18 and 35 (b) 15 and 37 (c) 30 and 415
(d) 17 and 68 (e) 216 and 215 (f) 81 and 16

Answer: (a) 18 and 35

Factors of 18=1,2,3,6,9,18

Factors of 35=1,5,7,35

Common factor =1

Therefore, the given two numbers are co-prime.

(b)15 and 37

Factors of 15=1,3,5,15

Factors of 37=1,37

Common factors =1

Therefore, the given two numbers are co-prime.

Factors of 30=1,2,3,5,6,10,15,30

Factors of 415=1,5,83,415

Common factors =1,5

As these numbers have a common factor other than 1, the given two numbers are non co-prime.

(d) 17 and 68

Factors of 17=1,17

Factors of 68=1,2,4,17,34,68

Common factors =1,17

As these numbers have a common factor other than 1, the given two numbers are non co-prime.

(e) 216 and 215

Factors of 216=1,2,3,4,6,8,9,12,18,24,27,36,54,72,108,216

Factors of 215=1,5,43,215

Common factors =1

Therefore, the given two numbers are co-prime.

(f) 81 and 16

Factors of 81=1,3,9,27,81

Factors of 16=1,2,4,8,16

Common factors =1

Therefore, the given two numbers are co-prime.

Q6 A number is divisible by both 5 and 12. By which other numbers will that number be always divisible?

Answer: Factors of 5=1,5

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

As the common factor of these numbers is 1, the given two numbers are coprime and the number will also be divisible by their product, i.e. 60, and the factors of 60=1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60.

Q7 A number is divisible by 12. By what other numbers will that number be divisible?

Answer: Since the number is divisible by 12, it will also be divisible by its factors i.e., 1, 2, 3, 4, 6, 12. Clearly, 1, 2, 3, 4, and 6 are numbers other than 12 by which this number is also divisible.

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime Factorization

Q Write the prime factorizations of 16, 28, 38.

Answer: prime factorizations of

16= 2×2×2×2
28= 2×2×7
38= 2×19

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.5

Q1 Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) A number is divisible by 18 if it is divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-primes, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number exactly divides the sum of two numbers, it must exactly divide the two numbers separately

Answer: (a) False as 6 is divisible by 3, but not by 9.

(b) True, as 9 = 3 x 3 Therefore, if a number is divisible by 9, then it will also be divisible by 3.

(d) True as 9 x 10 = 90 Therefore, If a number is divisible by 9 and 10 both, then it will also be divisible by 90.

(e) False as 15 and 32 are co-primes and also composite.

(f) False as 12 is divisible by 4, but not by 8.

(g) True, as 8 = 2 x 4 Therefore if a number is divisible by 8, then it will also be divisible by 2 and 4.

(h) True as 2 divides 4 and 8 as well as 12. (4 + 8 = 12)

(i) False as 2 divides 12, but not divide 7 and 5.

Q2 Here are two different factor trees for 60. Write the missing numbers.
(a)

Answer:

(a)

(b)

Q3 Which factors are not included in the prime factorization of a composite number?

Answer: 1 is the factors which are not especially included in the prime factorization of a composite number. When the number is divisible by 2 co-prime numbers then they are divisible by their product.

When two given numbers are divisible by a number, the sum is also divisible by number. When 2 given numbers are divisible by number then its difference is divisible by that number.

Q4 Write the greatest 4-digit number and express it in terms of its prime factors.

Answer: Greatest four-digit number = 9999

9999 = 3 $\times$ 3 $\times$ 11 $\times$ 101

Q5 Write the smallest 5-digit number and express it in the form of its prime factors.

Answer: Smallest five-digit number = 10,000

10000 = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 5 $\times$ 5 $\times$ 5 $\times$ 5

Q6 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: Prime factors of 1729 are -7,13,19.
Ascending order =7<13<19.
Relation = 7,13,19 differ by 6

Q7 The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.

Answer: 2 $\times$ 3 $\times$ 4=24, which is divisible by 6

9 $\times$ 10 $\times$ 11=990, which is divisible by 6

20 $\times$ 21 $\times$ 22=9240, which is divisible by 6

Q8 The sum of two consecutive odd numbers is divisible by 4. Verify this statement with the help of some examples.

Answer: 3+5=8, which is divisible by 4

15+17=32, which is divisible by 4

19+21=40, which is divisible by 4

Q9 In which of the following expressions, prime factorization has been done?
(a) 24 = 2 $\times$ 3 $\times$ 4 (b) 56 = 7 $\times$ 2 $\times$ 2 $\times$ 2
(c) 70 = 2 $\times$ 5 $\times$ 7 (d) 54 = 2 $\times$ 3 $\times$ 9

Answer: in factorization, we don't write composite numbers. All the factors should be prime numbers in this method.

(a) 24 = 2 $\times$ 3 $\times$ 4

As we know that 4 is a composite number.
Hence This is not a prime factorization.

b) 56 = 7 $\times$ 2 $\times$ 2 $\times$ 2
In this factorization, all the factors of 56 are prime numbers. There is no composite number.
Hence This is prime factorization.

c) 70 = 2 $\times$ 5 $\times$ 7
In this factorization, all the factors of 70 are prime numbers. There is not a composite number in this process of factorization.
Hence This is prime factorization.

d) 54 = 2 $\times$ 3 $\times$ 9
In this factorization, 54 is written as the product of 2, 3 and 9. In this factorization, 2 and 3 are prime numbers, but 9 is a composite number.
Hence This is not a prime factorization.

Q10 Determine if 25110 is divisible by 45.
[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9].

Answer: 45 = 5 $\times$ 9

Factors of 5 = 1, 5

Factors of 9 = 1. 3, 9

Therefore, 5 and 9 are co-prime numbers.

Since the last digit of 25110 is 0. it is divisible by 5.

Sum of the digits of 25110 = 2 + 5 + 1 +1 + 0 = 9

As the sum of the digits of 25110 is divisible by 9, therefore. 25110 is divisible by 9.

Since the number is divisible by 5 and 9 both, it is divisible by 45.

Q12 I am the smallest number, having four different prime factors. Can you find me?

Answer: Since it is the smallest number of such type, it will be the product of 4 smallest prime numbers. that is

2 $\times$ 3 $\times$ 5 $\times$ 7 = 210

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Highest Common Factor

Q Find the HCF of the following:
(i) 24 and 36 (ii) 15, 25 and 30
(iii) 8 and 12 (iv) 12, 16 and 28

Answer: (i) 24 and 36

24 = 2 $\times$ 2 $\times$ 2 $\times$ 3

36 = 2 $\times$ 2 $\times$ 3 $\times$ 3

HCF = 2 $\times$ 2 $\times$ 3 = 12

(ii) 15, 25 and 30

15 = 3 $\times$ 5

25 = 5 $\times$ 5

30 = 2 $\times$ 3 $\times$ 5

HCF = 5

(iii) 8 and 12

8 = 2 $\times$ 2 $\times$ 2

12 = 2 $\times$ 2 $\times$ 3

HCF = 2 $\times$ 2 = 4

(iv) 12, 16 and 28

12 = 2 $\times$ 2 $\times$ 3

16 = 2 $\times$ 2 $\times$ 2 $\times$ 2

28 = 2 $\times$ 2 $\times$ 7

HCF = 2 $\times$ 2 = 4

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.6

Q1 Find the HCF of the following numbers :
(a) 18, 48 (b) 30, 42
(c) 18, 60 (d) 27, 63
(e) 36, 84 (f) 34, 102
(g) 70, 105, 175 (h) 91, 112, 49
(i) 18, 54, 81 (j) 12, 45, 75

Answer: (a) 18, 48

18 = 2 x 3 x 3

48 = 2 x 2 x 2 x 2 x 3

HCF = 2 x 3 =6

(b) 30, 42

30 = 2 x 3 x 5

42 = 2 x 3 x 7

HCF = 2 x 3 =6

18 = 2 x 3 x 3

60 = 2 x 2 x 3 x 5

HCF = 2 x 3 = 6

(d) 27, 63

27 = 3 x 3 x 3

63 = 3 x 3 x 7

HCF = 3 x 3 = 9

(e) 36, 84

36 = 2 x 3 x 3 x 3

84 = 2 x 2 x 3 x 7

HCF = 2 x 2 x 3 = 12

(f) 34, 102

34 = 2 x 17

102 = 2 x 3 x 17

HCF = 2 x 17 = 34

(g) 70, 105, 175

70 = 2 x 5 x 7

105 = 3 x 5 x 7

175 = 5 x 5 x 7

HCF = 5 x 7 = 35

(h)91, 112, 49

91 = 7 x 13

112 = 2 x 2 x 2 x 2 x 7

49 = 7 x 7

HCF = 7

(i) 18, 54, 81

18 = 2 x 3 x 3

54= 2 x 3 x 3 x 3

81 = 3 x 3 x 3 x 3

HCF = 3 x 3 = 9

(j) 12, 45, 75

12 = 2 x 2 x 3

45 = 3 x 3 x 5

75 = 3 x 5 x 5

HCF = 2

Q2 What is the HCF of two consecutive
(a) numbers? (b) even numbers? (c) odd numbers?

Answer: (a) 1 e.g., HCF of 2 and 3 is 1.

(b) 2 e.g., HCF of 2 and 4 is 2.

Q3 HCF of co-prime numbers 4 and 15 was found as follows by factorization : 4 = 2 × 2 and 15 = 3 × 5 since there is no common prime factor, so HCF of 4 and 15 is 0. Is the answer correct? If not, what is the correct HCF?

Answer: No, the answer is not correct. 1 is the correct HCF.

4 = 2 $\times$ 2 $\times$ 1

15 = 3 $\times$ 5 $\times$ 1

1 is common in both so HCF is 1.

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.7

Q1 Renu purchases two bags of fertilizer of weights 75 kg and 69 kg. Find the maximum value of weight which can measure the weight of the fertilizer exact number of times.

Answer: Weight of the two bags = 75 kg and 69 kg

Maximum weight :

HCF (75, 69)

75 = 3 $\times$ 5 $\times$ 5

69 = 3 $\times$ 23

HCF = 3

Hence the maximum value of weight which can measure the weight of the fertilizer exact number of times is 3.

Q2 Three boys step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. What is the minimum distance each should cover so that all can cover the distance in complete steps?

Answer: Step measure of 1 st Boy =63cm

Step measure of 2 nd Boy =70cm

Step measure of 3 rd Boy =77cm

LCM of 63,70,77 :

Q3 The length, breadth and height of a room are 825 cm, 675 cm and 450 cm respectively. Find the longest tape which can measure the three dimensions of the room exactly.

Answer: Length =825cm=3 $\times$ 5 $\times$ 5 $\times$ 11

Breadth =675cm=3 $\times$ 3 $\times$ 3 $\times$ 5 $\times$ 5

Height =450cm=2 $\times$ 3 $\times$ 3 $\times$ 5 $\times$ 5

Longest tape = HCF of 825,675, and 450= 3 $\times$ 5 $\times$ 5

=75cm

Therefore, the longest tape is 75cm.

Q4 Determine the smallest 3-digit number which is exactly divisible by 6, 8 and 12.

Answer: Smallest number = LCM of 6, 8, 12

We have to find the smallest 3-digit multiple of 24.

It can be seen that

24 $\times$ 4 = 96 and

24 $\times$ 5 = 120.

Hence, the smallest 3-digit number which is exactly divisible by 6, 8, and 12 is 120.

Q5 Determine the greatest 3-digit number exactly divisible by 8, 10 and 12.

Answer: CM of 8, 10 and 12

LCM = 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 5 = 120

We have to find the greatest 3-digit multiple of 120.

It can be seen that

120 $\times$ 8 = 960 and

120 $\times$ 9 = 1080.

Hence, the greatest 3-digit number exactly divisible by 8, 10, and 12 is 960.

Q6 The traffic lights at three different road crossings change after every 48 seconds, 72 seconds and 108 seconds respectively. If they change simultaneously at 7 a.m., at what time will they change simultaneously again?

Answer: The time period after which these lights will change = LCM of 48, 72 and 108

LCM=2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 3 $\times$ 3 = 432

They will change together after every 432 seconds i.e., 7 min 12 seconds.

Hence, they will change simultaneously at 7:07:12 am.

Q7 Three tankers contain 403 litres, 434 litres and 465 litres of diesel respectively. Find the maximum capacity of a container that can measure the diesel of the three containers exact number of times.

Answer: Maximum capacity of the required tanker =HCF of 403,434,465

403=13 $\times$ 31

434=2 $\times$ 7 $\times$ 31

465=3 $\times$ 5 $\times$ 31

HCF=31

Hence the maximum capacity of a container that can measure the diesel of the three containers the exact number of times is 31.

Q8 Find the least number which when divided by 6, 15 and 18 leave remainder 5 in each case.

Answer: LCM of 6, 15, 18

LCM = 2 $\times$ 2 $\times$ 3 $\times$ 5 = 90

Required number = 90 + 5 = 95

Q9 Find the smallest 4-digit number which is divisible by 18, 24 and 32.

Answer: LCM of 18, 24 and 32

LCM = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3 $\times$ 3

We have to find the smallest 4-digit multiple of 288.

It can be observed that 288 $\times$ 3 = 864 and 288 $\times$ 4 = 1152.

Therefore, the smallest 4-digit number which is divisible by 18, 24, and 32 is 1152.

Q10 Find the LCM of the following numbers :
(a) 9 and 4 (b) 12 and 5 (c) 6 and 5 (d) 15 and 4
Observe a common property in the obtained LCMs. Is LCM the product of two numbers in each case?

Answer: (a) LCM = 2 $\times$ 2 $\times$ 3 $\times$ 3 = 36

(b) LCM = 2 $\times$ 2 $\times$ 3 $\times$ 5 = 60

(d) LCM = 2 $\times$ 3 $\times$ 5 = 30

Yes, it can be observed that in each case, the LCM of the given numbers is the product of these numbers.

When two numbers are co-prime, their LCM is the product of those numbers. Also, in each case, LCM is a multiple of 3.

Q11 Find the LCM of the following numbers in which one number is the factor of the other.
(a) 5, 20 (b) 6, 18 (c) 12, 48 (d) 9, 45
What do you observe in the results obtained?

Answer: (a) LCM = 2 $\times$ 2 $\times$ 5 = 20

(b) LCM = 2 $\times$ 3 $\times$ 3 = 18

(d) LCM = 3 $\times$ 3 $\times$ 5 = 45

Yes, it can be observed that in each case, the LCM of the given numbers is the larger number. When one number is a factor of the other number, their LCM will be the larger number.

Chapter No.	Chapter Name
Chapter 1	NCERT Solutions for class 6 maths chapter 1 Knowing Our Numbers
Chapter 2	NCERT solutions for class 6 maths chapter 2 Whole Numbers
Chapter 3	NCERT solutions for class 6 maths chapter 3 Playing with Numbers
Chapter 4	NCERT Solutions for class 6 maths chapter 4 Basic Geometrical Ideas
Chapter 5	NCERT solutions for class 6 maths chapter 5 Understanding Elementary Shapes
Chapter 6	NCERT solutions for class 6 maths chapter 6 Integers
Chapter 7	NCERT Solutions for class 6 maths chapter 7 Fractions
Chapter 8	NCERT solutions for class 6 maths chapter 8 Decimals
Chapter 9	NCERT solutions for class 6 maths chapter 9 Data Handling
Chapter 10	NCERT solutions for class 6 maths chapter 10 Mensuration
Chapter 11	NCERT Solutions for class 6 maths chapter 11 Algebra
Chapter 12	NCERT solutions for class 6 maths chapter 12 Ratio and Proportion
Chapter 13	NCERT solutions for class 6 maths chapter 13 Symmetry
Chapter 14	NCERT Solutions for class 6 maths chapter 14 Practical Geometry

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NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.1

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime and Composite Numbers

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.2

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.3

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Common Factors and Common Multiples

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.4

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime Factorization

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.5

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Highest Common Factor

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.6

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.7

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NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.1

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime and Composite Numbers

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.2

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.3

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Common Factors and Common Multiples

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.4

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Prime Factorization

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.5

NCERT solutions for class 6 maths chapter 3 Playing with numbers Topic: Highest Common Factor

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.6

NCERT solutions for class 6 maths chapter 3 Playing with numbers Exercise: 3.7

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