RS Aggarwal Solutions for Class 6 Maths Chapter 2 Factors and Multiples

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rs aggarwal solution class 6 maths chapter 2
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rs aggarwal solution class 6 maths chapter 2
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rs aggarwal solution class 6 maths chapter 2
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rs aggarwal solution class 6 maths chapter 2

 

Access answers to Chapter 2 – Factors and Multiples

Exercise 2A PAGE No: 25

1. Define: (i) factor (ii) multiple. Give five examples of each.

Solution

Factor: The exact divisor of a number is known as a factor

Multiple: A number obtained by multiplying it by natural number is known as multiple

Example: 1

We know that 14 = 1 × 14 = 2 × 7

Hence 1, 2, 7 and 14 are the factors of 14

∴ 14 is a multiple of 1, 2, 7 and 14

Example: 2

We know that 8 = 1 × 8 = 2 × 4

8 = 4 × 2

Hence 1, 2, 4 and 8 are the factors of 8

∴ 8 is a multiple of 1, 2, 4 and 8

2. Write down all the factors of

(i) 20 (ii) 36 (iii) 60 (iv) 75

Solution

(i) 20

20 = 1× 20

20 = 4 × 5

20 = 10 × 2

Hence the factors are 1, 2, 4, 5, 10 and 20

(ii) 36

36 = 1 × 36

36 = 6 ×6

36 = 9 × 4

36 = 12 × 3

36 = 18 × 2

Hence the factors are 1, 2, 3, 4, 6, 9,12,18 and 36

(iii) 60

60 = 1 × 60

60 = 10 × 6

60 = 30 × 2

60 = 15 × 4

60 = 12 × 5

60 = 3 × 20

Hence the factors are 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30 and 60

(iv) 75

75 = 1 × 75

75 = 3 × 25

75 = 5 × 15

Hence the factors are 1, 3, 5, 15, 25 and 75

3. Write the first five multiples of each of the following numbers:

(i) 17 (ii) 23 (iii) 65 (iv) 70

Solution

(i) 17

17× 1 = 17

17 × 2 = 14

17 × 3 = 51

17 × 4 = 68

17 × 5 = 85

∴ The first five multiples of 17 are 17, 34, 51, 68 and 85

(ii) 23

23 × 1 = 23

23 × 2 = 46

23 × 3 = 69

23 × 4 = 92

23 × 5 = 115

∴ The first five multiples of 23 are 23, 46, 69, 92 and 115

(iii) 65

65 × 1 = 65

65 × 2 = 130

65 × 3 = 195

65 × 4 = 260

65 × 5 = 325

∴ The first five multiples of 65 are 65, 130, 195, 260 and 325

(iv) 70

70 × 1 = 70

70 × 2 = 140

70 × 3 = 210

70 × 4 = 280

70 × 5 = 350

∴ The first five factors are 70, 140, 210 280 and 350

4. Which of the following numbers are even and which are odd?

(i) 32 (ii) 37 (iii) 50 (iv) 58

(v) 69 (vi) 144 (vii) 321 (viii) 253

Solutions

(i) 32

32 is a multiple of 2

Hence it’s an even number

(ii) 37

37 is not a multiple of 2

Hence it’s an odd number

(iii) 50

50 is a multiple of 2

Hence it’s an even number

(iv) 58

58 is a multiple of 2

Hence it’s an even number

(v) 69

69 is not a multiple of 2

Hence it’s an odd number

(vi) 144

144 is a multiple of 2

Hence it’s an even number

(vii) 321

321 is not a multiple of 2

Hence it’s an odd number

(viii) 253

253 is not a multiple of 2

Hence it’s an odd number

5. What are prime numbers? Give ten examples

Solutions

A number which has only two factors namely 1 and itself is called as prime numbers

Examples: 2, 3, 5, 7, 11, 13, 17, 19, 23 and 29 are prime numbers

6. Write all the prime numbers between

(i) 10 and 40 (ii) 80 and 100 (iii) 40 and 80 (iv) 30 and 40

Solutions

(i) 11, 13, 17, 19, 23, 29, 31 and 37 are all prime numbers between 10 and 40

(ii) 83, 89 and 97 are all prime numbers between 80 and 100

(iii) 41, 43, 47, 53, 59, 61, 67, 71, 73 and 79 are all prime numbers between 40 and 80

(iv) 31 and 37 are prime numbers between 30 and 40

7. (i) Write the smallest prime number.

(ii) List all even prime numbers

(iii) Write the smallest odd prime number

Solution

(i) 2 is the smallest prime number

(ii) 2 is only one even prime number

(iii) 3 is the smallest odd prime number

8. Find which of the following prime numbers are primes:

(i) 87 (ii) 89 (iii) 63 (iv) 91

Solutions

(i) 87

1, 3, 29 and 87 are the divisors of 87

Hence 87 has more than 2 factors

∴ 87 is not a prime number

(ii) 89

1 and 89 are the divisors of 89

∴ 89 is a prime number

(iii) 63

1, 3, 7, 9, 21 and 63 are the divisors of 63

Hence 63 has more than 2 factors

∴ 63 is not a prime number

(iv) 91

1, 7, 13 and 91 are the divisors of 91

Hence 91 has more than two factors

∴ 91 is not a prime number

9. Make a list of seven consecutive numbers, none of which is prime

Solution

The seven consecutive numbers are 90, 91, 91, 93, 94, 95 and 96 which are not prime numbers

10. (i) Is there any counting number having no factor at all?

(ii) Find all the numbers having exactly one factor.

(iii) Find numbers between 1 and 100 having exactly three factors.

Solution

  1. Since every number has two factors i.e. 1 and itself .Hence there are no counting number having no factor at all.
  2. There is only one number which has exactly 1 factor i.e. is 1
  3. 4 , 9 , 25 and 49 are the numbers between 1 and 100 having exactly three factors

Exercise 2B PAGE NO: 29

1. Test the divisibility of the following numbers by 2:

(i) 2650 (ii) 69435 (iii) 59628

(iv) 789403 (v) 357986 (vi) 367314

Solutions

A number is divisible by 2 only if its ones digit is 0, 2, 4, 6 and 8

(i) Since 0 is in one’s digits place in 2650

∴ It is divisible by 2

(ii) Since 0, 2, 4, 6 or 8 is not in ones digits place in 69435

∴ It is not divisible by 2

(iii) Since 8 is in ones digits place in 59628

∴ It is divisible by 2

(iv) Since 0, 2, 4, 6, or 8 is not in ones digits place in 789403

∴ It is not divisible by 2

(v) Since 6 is in ones digits place in 357986

∴ It is divisible by 2

(vi) Since 4 is in ones digits place in 367314

∴It is divisible by 2

2. Test the divisibility of the following numbers by 3

(i) 733 (ii) 10038 (iii) 20701

(iv) 524781 (v) 79124 (vi) 872645

Solutions

A number is divisible by 3 when sum of its digits is divisible by 3

(i) Sum of its digits = 7 + 3 + 3 = 13

Since 13 is not divisible by 3

∴ 733 is not divisible by 3

(ii) Sum of its digits = 1 + 0 + 0+ 3 +8 = 12

Since 12 is divisible by 3

∴ 10038 is divisible by 3

(iii) Sum of its digits = 2 + 0 + 7 + 0 + 1 = 10

Since 10 is not divisible by 3

∴ 20701 is not divisible by 3

(iv) Sum of its digits = 5 + 2 + 4 + 7 + 8 + 1 = 27

Since 27 is divisible by 3

∴ 524781 is divisible by 3

(v) Sum of its digits = 7+ 9 + 1 + 2 + 4 = 23

Since 23 is not divisible by 3

∴ 79124 is not divisible by 3

(vi) Sum of its digits 8 + 7 + 2 + 6 + 4 + 5 = 32

Since 32 is not divisible by 3

∴ 872645 is not divisible by 3

3. Test the divisibility of the following numbers by 4

(i) 618 (ii) 2314 (iii) 63712

(iv) 35056 (v) 946126 (vi) 810524

Solutions

A number is divisible by 4 if the digits in its ones and tens place are divisible by 4

  1. 618 is not divisible by 4 since the last two digits 18 is not divisible by 4
  2. 2314 is not divisible by 4 since last two digits is 14 is not divisible by 4
  3. 63712 is divisible by 4 since last two digits 12 is divisible by 4
  4. 35056 is divisible by 4 since last two digits 56 is divisible by 4
  5. 946126 is not divisible by 4 since last two digits 26 is not divisible by 4
  6. 810524 is divisible by 4 since last two digits 24 is divisible by 4

4. Test the divisibility of the following numbers by 5:

(i) 4965 (ii) 23590 (iii) 35208

(iv) 723405 (v) 124684 (vi) 438750

Solutions

If the ones digit of a number is 0 or 5 then the number is divisible by 5

(i) Since 5 is in ones place

∴ 4965 is divisible by 5

(ii) Since 0 is in ones place

∴ 23590 is divisible by 5

(iii) Since 8 is in ones place

∴ 35208 is not divisible by 5

(iv) Since 5 is in ones place

∴ 723405 is divisible by 5

(v) Since 4 is in ones place

∴ 124684 is not divisible by 5

(vi) Since 0 is in ones place

∴ 438750 is divisible by 5

5. Test the divisibility of the following numbers by 6:

(i) 2070 (ii) 46523 (iii) 71232

(iv) 934706 (v) 251780 (vi) 872536

Solutions

If the number is divisible by both 2 and 3 then the number is divisible by 6

(i) Divisibility by 2 = Since 0 is in ones place. It is divisible by 2

Divisibility by 3 = Since the sum of digits = 2 + 0 + 7 + 0 = 9

9 is divisible by 3

∴ 2070 is divisible by 6

(ii) Divisibility by 2 = Since 3 is in ones place it is not divisible by 2

∴ 46523 is not divisible neither by 2 or 6

(iii) Divisibility by 2 = Since 2 is in ones place it is divisible by 2

Divisibility by 3 = The sum of its digits = 7 + 1 + 2 + 3 + 2 = 15

15 is divisible by 3

∴ 71232 is divisible by 6 as it is divisible by 2 and 3

(iv) Divisibility by 2 = since sum of digits = 9 + 3 + 4 + 7 + 0 + 6 = 29

29 is not divisible by 3

∴ 934706 is not divisible by 6

(v) Divisibility by 3 = since sum of digits = 2 + 5 + 1 + 7 + 8 + 0 = 23

23 is not divisible by 3

∴ 251780 is not divisible by 6

(vi) Divisibility by 3 = since sum of digits = 8 + 7 + 2 + 5 + 3 +6 = 31

31 is not divisible by 3

∴ 872536 is not divisible by 6

6. Test the divisibility of the following numbers by 7:

(i) 826 (ii) 117 (iii) 2345

(iv) 6021 (v) 14126 (vi)25368

Solutions

To find out the number is divisible by 7, the last digit of the number is doubled and subtracted from the number formed by remaining digits. If the difference is multiple by 7 then its divisible by 7

(i) 826

82 – 2 × 6 = 70

70 is a multiple of 7

∴ 826 is divisible by 7

(ii) 117

11 – 2 × 7 = -3

-3 is not multiple of 7

∴ 117 is not divisible by 7

(iii) 2345

234 – 2 × 5 = 224

224 is a multiple of 7

∴ 2345 is divisible by 7

(iv) 6021

602 – 2 × 1 = 600

600 is not multiple of 7

∴ 6021 is not divisible by 7

(v) 14126

1412 – 2 × 6 = 1400

1400 is a multiple of 7

∴ 14126 is divisible by 7

(vi) 25368

2536 – 2 × 8 = 2520

2520 is a multiple of 7

∴ 25368 is divisible by 7

7. Test the divisibility of the following numbers by 8:

(i) 9364 (ii) 2138 (iii) 36792

(iv) 901674 (v) 136976 (vi) 1790184

Solutions

If the number formed by the last three digits is divisible by 8 then the given number is divisible by 8

(i) 9364

364 is not divisible by 8

Hence 9364 is not divisible by 8

(ii) 2138

138 is not divisible by 8

Hence 2138 is not divisible by 8

(iii) 36792

792 is divisible by 8

Hence 36792 is divisible by 8

(iv) 901674

674 is not divisible by 8

Hence 901674 is not divisible by 8

(v) 136976

976 is divisible by 8

Hence 136976 is divisible by 8

(vi) 1790184

184 is divisible by 8

Hence 1790184 is divisible by 8

8. Test the divisibility of the following numbers by 9:

(i) 2358 (ii) 3333 (iii) 98712

(iv) 257106 (v) 647514 (vi) 326999

Solutions

If the sum of its digits is divisible by 9 then the given number is divisible by 9

(i) 2358

Sum of its digits = 2 + 3 + 5 + 8 = 18

18 is divisible by 9

Hence 2358 is divisible by 9

(ii) 3333

Sum of its digits = 3 + 3 + 3 + 3 = 12

12 is not divisible by 9

Hence 3333 is not divisible by 9

(iii) 98712

Sum of its digits = 9 + 8 + 7 + 1 + 2 = 27

27 is divisible by 9

Hence 98712 is divisible by 9

(iv) 257106

Sum of its digits = 2 + 5 + 7 + 1 + 6 = 21

21 is not divisible by 9

Hence 257106 is not divisible by 9

(v) 647514

Sum of its digits = 6 + 4 + 7 + 5 + 1 + 4 = 27

27 is divisible by 9

Hence 647514 is divisible by 9

(vi) 326999

Sum of its digits = 3 + 2 + 6 + 9 + 9 + 9 = 38

38 is not divisible by 9

Hence 326999 is not divisible by 9

9. Test the divisibility of the following numbers by 10:

(i) 5790 (ii) 63215 (iii) 55555

Solutions

If the ones digit of a number is 0 then the given number is divisible by 10

(i) 5790

Here the ones digit is 0

∴ 5790 is divisible by 10

(ii) 63215

Here the ones digit is 5

∴ 63215 is not divisible by 10

(iii) 55555

Here the ones digit is 5

∴ 55555 is not divisible by 10

10. Test the divisibility of the following numbers by 11:

(i) 4334 (ii) 83721 (iii) 66311

(iv) 137269 (v) 901351 (vi) 8790322

Solutions

If the difference of the sum of its digits at odd places and sum of its digits at even places is either 0 or multiple of 11 then the given number is divisible by 11

(i) 4334

Sum of the digits at odd places = 4+ 3 = 7

Sum of the digits at even places = 3 + 4 = 7

Difference of the two sums = 7 – 7 = 0

∴ 4334 is divisible by 11

(ii) 83721

Sum of the digits at odd places = 1 + 7 + 8 =16

Sum of the digits at even places = 2 + 3 = 5

Difference of the two sums = 16 – 5 = 11

∴ 83721 is divisible by 11

(iii) 66311

Sum of the digits at odd places = 1 + 3 + 6 = 10

Sum of the digits at even places = 1 + 6 = 7

Difference of the two sums = 10 – 7 = 3

∴ 66311 is not divisible by 11

(iv) 137269

Sum of the digits at odd places = 9 + 2 + 3 = 14

Sum of the digits at even places = 6 + 7 + 1 = 14

Difference of the two sums = 14 – 14 = 0

∴ 137269 is divisible by 11

(v) 901351

Sum of the digits at odd places = 0 + 3 + 1 = 4

Sum of the digits at even places = 9 + 1 + 5 =15

Difference of the two sums = 4 – 15 = -11

∴ 901351 is divisible by 11

(vi) 8790322

Sum of the digits at odd places = 2 + 3 + 9 + 8 = 22

Sum of the digits at even places = 2 + 0 + 7 = 9

Difference of the two sums = 22-9 = 13

∴ 8790322 is not divisible by 11


Exercise 2C PAGE NO: 32

Give the prime factorization of each of the following numbers:

1. 12

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-1

∴ 12 = 2 × 2 × 3

= 22 × 3

The prime factorization of 12 = 22 × 3

2. 18

Solution

We have:

RS Aggarwal solutions For Class 6 Chapter 2 Exercise 2C-2

∴ 18 = 2 × 3 × 3

= 2 × 32

The prime factorization of 18 = 2 × 32

3. 48

Solution

We have:

Rs Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-3

∴ 48 = 2 × 2 × 2 × 2 ×3

= 24 × 3

The prime factorization of 48 = 24 × 3

4. 56

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-4

∴ 56 = 2 × 2 × 2 × 7

= 23 × 7

The prime factorization of 56 = 23 × 7

5. 90

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-5

∴ 90 = 2 × 3 × 3 × 5

= 2 × 32 × 5

The prime factorization of 90 = 2 × 32 × 5

6. 136

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-6

∴ 136 = 2 × 2 × 2 × 17

= 23 × 17

The prime factorization of 136 = 23 × 17

7. 252

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-7

∴ 252 = 2 × 2 × 3 × 3 × 7

= 22 × 32 × 7

The prime factorization of 252 = 22 × 32 × 7

8. 420

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-8

∴ 420 = 2 × 2 × 3 × 7 × 5

The prime factorization of 420 = 22 × 3 × 5 × 7

9.637

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-9

∴ 637 = 7 × 7 × 13

= 72 × 13

The prime factorization of 637 = 72 × 13

10. 945

Solution

We have:

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2C-10

∴ 945 = 3 × 3 × 3 × 5 × 7

= 33 × 5 × 7

The prime factorization of 945 = 33 × 5 × 7


Exercise 2D PAGE NO: 36

Find the HCF of the numbers in each of the following, using the prime factorization method:

1. 84, 98

Solution

RS Aggarwal Solutions For class 6 chapter 2 Exercise 2D-1

RS Aggarwal solutions for class 6 chapter 2 Exercise 2D-2

The prime factors of 84 = 2 × 2 × 3 × 7 = 22 × 3 × 7

The prime factors of 98 = 2 × 7 × 7 = 2 × 72

∴ HCF = 2× 7 = 14

2. 170, 238

Solution

RS Aggarwal Solutions for class 6 chapter 2 Exercise 2D-3


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-4

The prime factors of 170 = 2 × 5 × 17

The prime factors of 238 = 2 × 7 × 17

∴ HCF = 2 × 17 = 34

3. 504,980

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise  2D-5

The prime factors of 504 = 23 × 32 × 7

RS  Aggarwal Solutions for Class 6 Chapter 2 Exercise 2D-6

The prime factors of 980 = 22 × 5 × 72

∴ HCF = 22 × 7 = 28

  1. 72, 108, 180

Solution

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-7


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-8

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D- 9

The prime factors of 72 = 23 × 32

The prime factors of 108 = 23 ×3

The prime factors of 180 = 22 × 32 ×5

∴ HCF = 22 × 32 = 36

  1. 84, 120, 138

Solution

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-10


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D -11


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-12

The prime factors of 84 = 22 × 3 ×7

The prime factors of 120 = 23 × 3 × 5

The prime factors of 138 = 2 × 3 × 23

∴ HCF = 2 × 3 = 6

  1. 106,159,371

Solution

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-13

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-14

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-15

The prime factors of 106 = 2 × 53

The prime factors of 159 = 3 × 53

The prime factors of 371 = 7 × 53

∴ HCF = 53

  1. 272, 425

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-16

RS Aggarwal solutions For Class 6 Chapter 2 Exercise 2D-17

The prime factors of 272 = 24 × 17

The prime factors of 425 = 52 × 17

∴ HCF = 17

8.144, 252,630

Solution


RS Aggarwal solutions For Class 6 Chapter 2 Exercise 2D-18

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-19

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-20

The prime factors of 144 = 24 × 32

The prime factors of 252 = 22 × 32 × 7

The prime factors of 630 = 2 × 32 × 5

∴ HCF = 18

9. 1197, 5320, 4389

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-21


RS Aggarwal olutions For Class 6 Chapter 2 Exercise 2D-22


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-23

The prime factors of 1197 = 32 × 7 × 19

The prime factors of 5320 = 23 × 5 × 7 × 19

The prime factors of 4389 = 3× 7 × 11 × 19

∴ HCF = 133

Find the HCF of the numbers in each of the following, using the division method:

10. 58, 70

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-24

∴ The common factor of 58, 70 =2

11. 399, 437

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-25

∴ The Common factor of 399, 437 = 19

12. 1045, 1520

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-26

∴ The HCF of 1045, 1520 = 95

13. 1965, 2096

Solution

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-27

∴ Common factor of 1965, 2096 = 131

14. 2241, 2324

Solution


RS Aggarwal solutions For Class 6 Chapter 2 Exercise 2D-28

∴ The Common factor of 2241, 2324 = 83

15. 658, 940, 1128

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-29

∴ Common factor of 658, 940, 1128 = 94

  1. 754, 1508, 1972

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-30


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D-31

∴ Common factor of 754, 1508, 1972 = 58

17. 391, 425, 527

Solution

RS Aggarwal Solutions for Class 6 Chapter 2 Exercise 2D - 32

∴ Common factor of 391, 425, 527 = 17

18. 1794, 2346, 4761

Solution

RS Aggarwal Solutions for Class 6 Chapter 2 Exercise 2D- 33

∴ Common factor of 1794, 2346, 4761 is 69

Show that the following primes are co-primes:

19. 59, 97

Solution

59 = 59× 1

97 = 97 × 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

20. 161,192

Solution

161 = 7 × 23

192 = 2 × 2 × 2 × 2 × 2 × 2 ×3

= 26 ×3 ×1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

21.343, 432

Solution

343 = 7× 7 × 7

= 73 × 1

432 = 2 × 2 × 2 × 2 × 3 × 3 × 3

= 24 × 33 × 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

22. 512, 945

Solution

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2

= 29

945 = 3 × 3 × 3 × 5 × 7

= 33 × 5 × 7

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

23. 385, 621

Solution

385 = 5 × 7 × 11 × 1

621 = 3 × 3 × 3 × 23 × 1

= 33 × 23

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

24. 847, 1014

Solution

847 = 7 × 11 × 11 × 1

= 7 × 112 × 1

1014 = 2 × 3 × 13 × 13 × 1

= 2 × 3 × 132 × 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

25. Find the greatest number which divides 615 and 963, leaving the remainder 6 in each case

Solution

Since the remainder is 6 we have to find the number which exactly divides (615 -6) and (963 -6)

Required number = HCF of 609 and 957

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2D- 34

Hence, 87 is the greatest number which divides 615 and 963 leaving the remainder 6


Exercise 2E PAGE NO: 40

1. 42, 63

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E- 1


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-2

42 = 2 × 3 × 7

63 = 3 × 3× 7

= 32 × 7

∴ LCM of 42 and 63 = 2 × 32 × 7

= 2 × 9 × 7

= 18 × 7

= 126

2. 60, 75

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-3


RS Aggarwal Solutiuons For Class 6 Chapter 2 Exercise 2E-4

60 = 2 c 2 × 3 × 5

= 22 × 3 × 5

75 = 3 × 5 × 5

= 3 × 52

∴ LCM of 60 and 75 = 22 × 3 × 52

= 4 × 3 × 25

= 12 × 25

= 300

3. 12, 18, 20

Solutions


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E- 5


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-6


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-7

12 = 2 × 2 × 3 = 22 × 3

18 = 2 × 3 × 3 = 2 × 32

20 = 2 × 2 × 5 = 22 × 5

∴ LCM of 12, 18, and 20 = 22 × 32 × 5

= 4 × 9 × 5

= 180

4. 36, 60, 72

Solutions


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E- 8


RS Aggarwal soluions For Class 6 Chapter 2 Exercise 2E-9


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E- 10

36 = 2 × 2 × 3 × 3 = 2 2 × 32

60 = 2 × 2 × 3 × 5 = 22 × 3 × 5

72 = 2 × 2 × 2 × 3 × 3 = 23 × 32

∴ LCM of 36, 60 and 72 = 23 × 32 × 5

= 8 × 9 × 5

= 72 × 5

= 360

5. 36, 40, 126

Solutions


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-11


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E- 12


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-13

36 = 2 × 2 × 3 × 3 = 22 × 32

40 = 2 × 2 × 2 × 5 = 23 × 5

126 = 2 × 32 × 7

∴ LCM of 36, 40 and 126 = 23 × 32 × 5 × 7

= 8 × 9 × 5 × 7

= 72 × 35

= 2520

6. 16, 28, 40, 77

Solution


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-14

∴ LCM of given numbers = 2 × 2 × 2 × 7 × 2 × 5 × 11

= 24 × 5 × 7 × 11

= 16 × 35 × 11

= 6160

7. 28, 36, 45, 60

Solutions


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-15

∴ LCM of given numbers = 2 × 2 × 3 × 3 × 5 × 7

= 22 × 32 × 5 × 7

= 4 × 9 × 5 × 7

= 36 × 35

= 1260

8. 144, 180, 384

Solutions


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-16

∴ LCM of given numbers = 2 × 2 × 2 × 2 × 3 × 3 × 5 × 8

= 24 × 32 × 5 × 8

= 16 × 9 × 5 × 8

= 16 × 45 × 8

= 5760

9. 48, 64, 72, 96, 108

Solution

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-17

∴ LCM of given numbers = 2 × 2 × 2 × 2 × 2 × 3 × 3 × 2 × 3

= 26 × 33

= 64 × 27

= 1728

Find the HCF and LCM of

1. 117, 221

Solution

To find the HCF of 117 and 221

RS Aggarwal Solutions For Class 6  Chapter 2 Exercise 2E-18

∴ HCF of 117and 221 = 13

Since, LCM = product of numbers/HCF

=117 × 221/13

= 9 × 221

= 1989

∴ LCM = 1989 and HCF = 13

2. 234,572

Solution

To find the HCF of 234 and 572

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-19

∴ HCF of 234 and 572 = 26

Since, LCM = Product of numbers/ HCF

= 234 × 572/ 26

= 9 × 572

= 5148

∴ LCM = 5148 and HCF = 26

3. 693, 1078

Solution

To find the HCF of 693 and 1078


RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-20

∴ HCF of 693 and 1078 = 77

Since, LCM = Product of numbers/ HCF

= 693 × 1078 / 77

= 9 × 1078

= 9702

∴ LCM = 9702 and HCF = 77

4. 145,232

Solution

To find the HCF of 693 and 1078

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise  2E-21

∴ HCF of 145 and 232 = 29

Since, LCM = product of numbers / HCF

= 145 × 232 / 29

= 5 × 232

= 1160

∴ LCM = 1160 and HCF = 29

5. 861, 1353

Solution

To find HCF of 861 and 1353

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-22

∴ HCF of 861 and 1353 = 123

Since, LCM = product of numbers /HCF

= 861 × 1353 / 123

= 7 × 1353

= 9471

∴ LCM = 9471 and HCF = 123

6. 2923, 3239

Solution

To find HCF of 2923 and 3239

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2E-23

∴ HCF of 2923 and 3239 = 79

Since, LCM = product of numbers/ HCF

= 2923 × 3239 / 79

= 37 × 3239

= 119843

∴ LCM = 119843 and HCF = 79


Exercise 2F PAGE NO: 41

1. Which of the following numbers is divisible by 3?

(a) 24357806 (b) 35769812 (c) 83479560 (d) 3336433

Solution

83479560

Since sum of its digits = 8 + 3 + 4 + 7 + 9 + 5 + 6 + 0 = 42

42 is divisible by 3

Option (c) is the correct answer

2. Which of the following numbers is divisible by 9?

(a) 8576901 (b) 96345210 (c) 67594310 (d) none of these

Solution

8576901

Since sum of its digits = 8 + 5 + 7 + 6 + 9 + 0 + 1 = 36

36 is divisible by 9

Option (a) is the correct answer

3. Which of the following numbers is divisible by 4?

(a) 78653234 (b) 98765042 (c) 24689602 (d) 87941032

Solution

87941032

Since the number formed by tens and ones digits is divisible by 4 i.e. 32

32 ÷ 4 = 8

Option (d) is the correct answer

4. Which of the following numbers is divisible by 8?

(a) 96354142 (b) 37450176 (c) 57064214 (d) none of these

Solution

37450176

Since the number formed by hundreds tens and one’s digits is divisible by 8 i.e. 176

176 ÷ 8 = 22

Option (b) is the correct answer

5. Which of the following numbers is divisible by 6?

(a) 8790432 (b) 98671402 (c) 85492014 (d) none of these

Solution

8790432

Since its one digit is divisible by 2 and

Sum of its digits = 8 + 7 + 9 + 0 + 4 + 3 + 2 = 33

33 is divisible by 3, hence divisible by 6

Option (a) is the correct answer

6. Which of the following numbers is divisible by 11?

(a) 3333333 (b) 1111111 (c) 22222222 (d) none of these

Solution

22222222

Since the difference of the sum of its odd places digits and of its even places digits is

(2 + 2 + 2 + 2) – (2 + 2 + 2 + 2) = 8 – 8 = 0

Hence divisible by 11

Option (c) is the correct answer

7. Which of the following is a prime number?

(a) 81 (b) 87 (c) 91 (d) 97

Solution

c) 97

Since 97 have no factors other than 1 and itself

Option (d) is the correct answer

8. Which of the following is a prime number?

(a) 117 (b) 171 (c) 179 (d) none of these

Solution

179

Since 179 have no factors other than 1 and itself

Option (c) is the correct answer

9. Which of the following is a prime number?

(a) 323 (b) 361 (c) 263 (d) none of these

Solution

a) 323 can be written as 17 × 19

Hence 323 is not a prime number

b) 361 can be written as 19 × 19

Hence 361 is not a prime number

c) 263 is a prime number

Option (c) is the correct answer

10 Which of the following are co-primes?

(a) 8, 12 (b) 9, 10 (c) 6, 8 (d) 15, 18

Solution

(a) 8, 12

Here they have a common factor 4

Hence 8, 12 are not co- primes

(b) 9, 10

Here they don’t have a common factor

Hence 9, 10 are co- primes

(c) 6, 8

Here they have a common factor 2

Hence 6, 8 are not co- primes

(d) 15, 18

Here they have a common factor 3

Hence 15, 18 are not co-primes

Option (b) is the correct answer

11. Which of the following is a composite number?

(a) 23 (b) 29 (c) 32 (d) none of these

Solution

(a) 23

Since it cannot be broken into factors

Hence 23 is not a composite number

(b) 29

Since it cannot be broken into factors

Hence 29 is not a composite number

(c) 32

Since it can be broken into factors i.e. 2 × 2 × 2 × 2 × 2

Hence 32 is a composite number

Option (c) is the correct answer

12. The HCF of 144 and 198 is

(a) 9 (b) 12 (c) 6 (d) 18

Solution

First factorize the two numbers

RS Agarwal Solutions For Class 6  Chapter 2 Exercise 2F-1
RS Aggarwal Solutions For Class 6 Chapter 2  Exercise 2F-2

Here 144 = 2 × 2 × 2 × 2 × 3 × 3

= 24 × 32

198 = 2 × 3 × 3 × 11

= 2 × 32 × 11

18 = 2 × 32 is the highest common factor

Option (d) is the correct answer

13. The HCF of 144, 180 and 192 is

(a) 12 (b) 16 (c) 18 (d) 8

Solution

Here 22 × 3 = 12

The factorization of 144, 180 and 192

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2F-3
RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2F-4
RS Aggarwal solutions for class 6 Chapter 2 Exercise 2F-5

Hence 144 = 2 × 2 × 2 × 2 × 3 × 3 = 24 × 32

180 = 2 × 2 × 3 × 3 × 5 = 22 × 32 × 5

192 = 2 × 2 × 2 × 2 × 2 × 2 × 3 = 26 × 3

Hence 22 × 3 is the highest common factor of all the three numbers

Option (a) is the correct answer

14. Which of the following are co-primes?

(a) 39, 91 (b) 161, 192 (c) 385, 462 (d) none of these

Solutiuon

a) 39, 91

Since 39, 91 have common factor 13

Hence 39, 91 are not co-primes

b) 161, 192

Since 161, 192 have no common factor other than 1 itself

Hence 161, 192 are co-primes

c) 385, 462

Since 385, 462 have common factors 7 and 11

Hence 385, 462 are not co-primes

Option (b) is the correct answer

15. 289/391 when reduced to the lowest terms is

(A)11/23 (b) 13/31 (c) 17/ 31 (d) 17/23

Solution

Dividing both numerator and denominator by the HCF of 289/391

RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2F-6
RS Aggarwal Solutions For Class 6 Chapter 2 Exercise 2F-7

Hence 289/391 = 17/23

17/23 is the correct answer

Option (d) is the correct answer

16. The greatest number which divides 134 and 167 leaving 2 as remainder in each case is

(A)14 (b) 17 (c) 19 (d) 33

Solution

Since we require 2 as the remainder we will subtract 2 from each of the numbers

167 – 2 = 165

134 – 2 = 132

Now any of the common factors of 165 and 132 will be the required divisor

On factorization

165 = 3 × 5 × 11

132 = 2 × 2 × 3 × 11

Their common factors are 11 and 3 = 11 × 3 = 33

Hence 33 is the required divisor

Option (d) is the correct answer


RS Aggarwal Solutions for Class 6 Maths Chapter 2 Factors and Multiples

Chapter 2 – Factors and Multiples includes 6 exercises. For each question the RS Aggarwal Solutions are solved in detail in every exercise. Let’s have a look at the topics mentioned in this chapter

  • Finding prime and composite numbers from 1 to 100
  • Divisibility tests for 2, 3, 4, 5, 6, 7, 8, 9, 10 and 11
  • General properties of divisibility
  • To find prime numbers between 100 and 400
  • Prime factorization
  • Lowest common multiple
  • Properties of HCF and LCM of given numbers

Also, access RS Aggarwal Solutions for Class 6 Chapter 2 Exercises

Exercise 2A

Exercise 2B

Exercise 2C

Exercise 2D

Exercise 2E

Exercise 2F

Chapter Brief of RS Aggarwal Solutions for Class 6 Maths Chapter 2 – Factors and Multiples

Factors and Multiples are both different things, but both involve multiplication. A factor of a number is an exact divisor of that number and a number is said to be a multiple of any of its factors. Factors and multiples are used in dividing something equally, factoring with money, comparing prices, understanding time and making calculations while travelling.