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

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

 

Solutions of NCERT 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

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

(b) 15  = 1 \times 15

           = 3  \times

           = 5  \times  3   

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

(c) 21 = 1 \times 21 

          =  3  \times  7

          =  7  \times  3     

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

(d) 27 = 1 \times 27 

           = 3 \times

           = 9 \times

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

(e) 12=1\times12

         =2\times6

         =3\times4

         =4\times3

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

          = 6 \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

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

(c) 9

   9 \times 1  =  9

   9 \times 2  = 18

   9 \times 3  = 27

   9 \times 4 =  36 

   9 \times 5 =  45 

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 \times4 = 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

 

Solutions of NCERT 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 .

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

(c) 24 = 19 +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 \times1 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.

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

(c) 53 = 3 + 19 + 31

(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

 

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

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

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.

(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.

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.

(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.

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

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.

 

Solutions of NCERT 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

 

(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

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. 

 

(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.

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

Write the prime factorizations of 16, 28, 38.

Answer:

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

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.

(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.

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\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

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.

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\times 5 \times 7 = 210

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

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

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

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

 

(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

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.

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

\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}

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 \times11

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

 Height =450cm=2 \times3 \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. 

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

Answer:

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.

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

Answer:

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.

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

\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.

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
\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

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

Answer:

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.

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

\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.

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

\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.
 

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