# NCERT Solutions for Class 6 Maths Chapter 3 Playing with Numbers

NCERT solutions for class 6 maths chapter 3 Playing with Numbers- The chapter starts with problems that introduce the concepts of multiples and divisors. As it progresses, multiples, divisors, and factors add fun to the chapter in the form of a game. CBSE NCERT solutions for class 6 maths chapter 3 Playing with Numbers is covering the solutions from each concept. The chapter is introduced to make mathematics interesting to class 6 students. In chapter 3 Playing with Numbers, students will learn about prime and composite numbers, tests for divisibility of numbers, common multiples, and common factors, prime factorization, some more divisibility rules, highest common factor (HCF), lowest common multiple (LCM). Solutions of NCERT for Class 6 Maths Chapter 3 Playing with Numbers are written very elegantly keeping step by step exam marking in the mind. In this particular chapter, there are a total of 7 exercises with a total of 55 questions. NCERT solutions for class 6 maths chapter 3 Playing with numbers will benefit you in scoring the maximum possible marks in mathematics. These solutions are covering all the 55 questions. NCERT solutions can be a good tool for your preparation if you make its full use. The exercises are listed below. Click on the link to jump to respective exercise.

Exercise 3.1

Exercise 3.2

Exercise 3.3

Exercise 3.4

Exercise 3.5

Exercise 3.6

Exercise 3.7

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

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

## Solutions of NCERT for class 6 maths chapter 3 Playing with numbers Exercise: 3.1

(a) 24 = 1 $\dpi{100} \times$ 24

= 2 $\dpi{100} \times$ 12

=  3 $\dpi{100} \times$ 8

= 4 $\dpi{100} \times$

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

(b) 15  = 1 $\dpi{100} \times$ 15

= 3  $\dpi{100} \times$

= 5  $\dpi{100} \times$  3

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

(c) 21 = 1 $\dpi{100} \times$ 21

=  3  $\dpi{100} \times$  7

=  7  $\dpi{100} \times$  3

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

(d) 27 = 1 $\dpi{100} \times$ 27

= 3 $\dpi{100} \times$

= 9 $\dpi{100} \times$

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

(e) 12=1$\dpi{100} \times$12

=2$\dpi{100} \times$6

=3$\dpi{100} \times$4

=4$\dpi{100} \times$3

Hence Factors of 12 are 1,2,3,4,6, and 12
(f) 20 = 1 $\dpi{100} \times$ 20

= 2 $\dpi{100} \times$ 10

= 4  $\dpi{100} \times$  5

= 5  $\dpi{100} \times$  4

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

(g) 18  = 1  $\dpi{100} \times$ 18

= 2  $\dpi{100} \times$  9

= 3  $\dpi{100} \times$ 6

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

(h) 23 = 1 $\dpi{100} \times$ 23

= 23 $\dpi{100} \times$ 1

Hence factors of 23 are 1 and 23.

(i) 36 = 1 $\dpi{100} \times$ 36

= 2  $\dpi{100} \times$ 18

= 3 $\dpi{100} \times$ 12

= 4 $\dpi{100} \times$

= 6 $\dpi{100} \times$

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

First Five multiple of

(a) 5

5 $\dpi{100} \times$ 1  =  5

5 $\dpi{100} \times$ 2  = 10

5 $\dpi{100} \times$ 3  = 15

5 $\dpi{100} \times$ 4 =  20

5 $\dpi{100} \times$ 5 =  25

(b) 8

8 $\dpi{100} \times$ 1  =  8

8 $\dpi{100} \times$ 2  = 16

8 $\dpi{100} \times$ 3  = 24

8 $\dpi{100} \times$ 4 =  32

8 $\dpi{100} \times$ 5 =  40

(c) 9

9 $\dpi{100} \times$ 1  =  9

9 $\dpi{100} \times$ 2  = 18

9 $\dpi{100} \times$ 3  = 27

9 $\dpi{100} \times$ 4 =  36

9 $\dpi{100} \times$ 5 =  45

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

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

some more numbers of this type are :

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

## Solutions of NCERT for class 6 maths chapter 3 Playing with numbers Exercise: 3.2

(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

(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 .

Prime Number having the same digit are:

17,71

37,73 and

79,97

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

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

(a) 44 = 37 +7

(b) 36 = 31 +5

(c) 24 = 19 +5

(d) 18 = 11 +7

three pairs of prime numbers whose difference is 2 are :

3,5

41,43 and

71,73

(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.

(c) 37 It has only two factors, 1 and 37. Therefore, it is a prime number.

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

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

(a) 21 = 3 + 7 + 11

(b) 31 = 5 + 7 + 19

(c) 53 = 3 + 19 + 31

(d) 61 = 11 + 19 + 31

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

(a) prime number

(b) Composite number

(c) Prime number, Composite number

(d) 2

(e) 4

(f) 2

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

 Number Divisible by 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 ...... ...... ...... ...... ...... ...... ...... ...... ......

 Number Divisible by 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

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.

(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.

(c) 4335

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.

(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.

(c) 7138965

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.

(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.

(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

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 $1\ and\ 3$.

## Solutions of NCERT for class 6 maths chapter 3 Playing with numbers Exercise: 3.4

(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

(c) 35 and 50

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

(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

(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

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

(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.

(c) 30 and 415

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.

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.

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

prime factorizations of

16= 2×2×2×2
28= 2×2×7
38= 2×19

## Solutions of NCERT for class 6 maths chapter 3 Playing with numbers Exercise: 3.5

(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.

(c) False as 30 is divisible by 3 and 6 both, but it is not divisible by 18.

(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.

(a)

(b)

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.

Greatest four-digit number = 9999

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

Smallest five-digit number = 10,000

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

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

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

3+5=8, which is divisible by 4

15+17=32, which is divisible by 4

19+21=40, which is divisible by 4

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

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

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$$\times$
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.

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.

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

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

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

(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 $\times$

25 = 5 $\times$

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 $\times$

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

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

HCF = 2 $\times$ 2 = 4

## Solutions of NCERT for class 6 maths chapter 3 Playing with numbers Exercise: 3.6

(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

(c)  18, 60

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

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

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

(c) 1 e.g., HCF of 3 and 5 is 1.

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

As

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

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.

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 :

$\begin{array}{|c|c|}\hline 2 & {63,70,77} \\ \hline 3 & {63,35,77} \\ \hline 3 & {21,35,77} \\ \hline 5 & {7,35,77} \\ \hline 7 & {7,7,11} \\ \hline 11 & {1,1} \\ \hline\end{array}$

$\begin{array}{l}{\text { L.C.M. of } 63,70 \text { and } 77=7 \times 9 \times 10 \times 11=6930 \mathrm{cm} .} \\ {\text { Therefore, the minimum distance is } 6930 \mathrm{cm} .}\end{array}$

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

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

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

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

=75cm

Therefore, the longest tape is 75cm.

Smallest number = LCM of 6, 8, 12

$\begin{array}{|c|c|c|}\hline 2 & {6,8,12} \\ \hline 2 & {3,4,6} \\ \hline 2 & {3,2,3} \\ \hline 3 & {3,1,3} \\ \hline 3 & {1,1,1} \\ \hline\end{array}$

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.

CM of 8, 10 and 12

$\begin{array}{|c|c|}\hline 2 & {8,10,12} \\ \hline 2 & {4,5,6} \\ \hline 2 & {2,5,3} \\ \hline 3 & {1,5,3} \\ \hline 5 & {1,5,1} \\ \hline\end{array}$

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.

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

$\begin{array}{|c|c|}\hline 2 & {48,72,108} \\ \hline 2 & {24,36,54} \\ \hline 2 & {12,18,27} \\ \hline 2 & {6,9,27} \\ \hline 3 & {1,3,9} \\ \hline 3 & {1,1,3} \\ \hline 3 & {1,1,1} \\ \hline\end{array}$

LCM=2 $\times$$\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.

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.

LCM of 6, 15, 18
$\begin{array}{|c|c|c|}\hline 2 & {6,15,18} \\ \hline 3 & {3,15,9} \\ \hline 3 & {1,5,3} \\ \hline 5 & {1,5,1} \\ \hline & {1,1,1} \\ \hline\end{array}$

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

Required number = 90 + 5 = 95

LCM of 18, 24 and 32

$\begin{array}{|c|c|}\hline 2 & {18,24,32} \\ \hline 2 & {9,12,16} \\ \hline 2 & {9,6,8} \\ \hline 2 & {9,3,4} \\ \hline 2 & {9,3,2} \\ \hline 3 & {9,3,1} \\ \hline 3 & {3,1,1} \\ \hline & {1,1,1} \\ \hline\end{array}$

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.

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

$\begin{array}{|c|c|}\hline 2 & {9,4} \\ \hline 2 & {9,2} \\ \hline 3 & {9,1} \\ \hline 3 & {3,1} \\ \hline & {1,1} \\ \hline\end{array}$

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

$\begin{array}{|c|c|}\hline 2 & {12,5} \\ \hline 2 & {6,5} \\ \hline 3 & {3,5} \\ \hline 5 & {1,5} \\ \hline & {1,1} \\ \hline\end{array}$

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

$\begin{array}{|c|c|}\hline 2 & {6,5} \\ \hline 3 & {3,5} \\ \hline 5 & {1,5} \\ \hline & {1,1} \\ \hline\end{array}$

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

$\begin{array}{|c|c|}\hline 2 & {15,4} \\ \hline 2 & {15,2} \\ \hline 3 & {15,1} \\ \hline 5 & {5,1} \\ \hline & {1} \\ \hline\end{array}$

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.

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

$\begin{array}{|c|c|}\hline 2 & {5,20} \\ \hline 2 & {5,10} \\ \hline 5 & {5,5} \\ \hline & {1,1} \\ \hline\end{array}$

(b) LCM = 2 $\times$$\times$ 3 = 18

$\begin{array}{|c|c|}\hline 2 & {6,18} \\ \hline 3 & {3,9} \\ \hline 3 & {1,3} \\ \hline & {1,1} \\ \hline\end{array}$

(c) LCM = 2 $\times$ 2 $\times$ 2 $\times$ 2 $\times$ 3 = 48

$\begin{array}{|c|c|}\hline 2 & {12,48} \\ \hline 2 & {6,24} \\ \hline 2 & {3,12} \\ \hline 2 & {3,12} \\ \hline 3 & {3,3} \\ \hline & {1,1} \\ \hline\end{array}$

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

$\begin{array}{|c|c|}\hline 3 & {9,45} \\ \hline 3 & {3,15} \\ \hline 5 & {1,5} \\ \hline & {1,1} \\ \hline\end{array}$

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.

## NCERT solutions for class 6 mathematics chapter-wise

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

## NCERT solutions for class 6 subject wise

 NCERT Solutions for class 6 maths Solutions of NCERT for class 6 science

## How to use NCERT solutions for class 6 maths chapter 3 Playing with Numbers

• Learn to differentiate between the multiples and factors.
• Go through some important concepts given in the NCERT textbook.
• Take a look through some examples to understand the pattern of solutions for a particular question.
• Once you are done with all the above-written points then you can directly come to the exercises given in the NCERT textbook.
• While doing the question of exercise you can take help of NCERT solutions for class 6 maths chapter 3 Playing with Numbers.