## Class 6 RS Aggarwal Chapter 2 – Factors and Multiples Solutions

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## 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 Ã— 14 = 2 Ã— 7

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

âˆ´ 14 is a multiple of 1, 2, 7 and 14

Example: 2

We know that 8 = 1 Ã— 8 = 2 Ã— 4

8 = 4 Ã— 2

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

âˆ´ 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 = 1Ã— 20

20 = 4 Ã— 5

20 = 10 Ã— 2

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

(ii) 36

36 = 1 Ã— 36

36 = 6 Ã—6

36 = 9 Ã— 4

36 = 12 Ã— 3

36 = 18 Ã— 2

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

(iii) 60

60 = 1 Ã— 60

60 = 10 Ã— 6

60 = 30 Ã— 2

60 = 15 Ã— 4

60 = 12 Ã— 5

60 = 3 Ã— 20

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

(iv) 75

75 = 1 Ã— 75

75 = 3 Ã— 25

75 = 5 Ã— 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

17Ã— 1 = 17

17 Ã— 2 = 14

17 Ã— 3 = 51

17 Ã— 4 = 68

17 Ã— 5 = 85

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

(ii) 23

23 Ã— 1 = 23

23 Ã— 2 = 46

23 Ã— 3 = 69

23 Ã— 4 = 92

23 Ã— 5 = 115

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

(iii) 65

65 Ã— 1 = 65

65 Ã— 2 = 130

65 Ã— 3 = 195

65 Ã— 4 = 260

65 Ã— 5 = 325

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

(iv) 70

70 Ã— 1 = 70

70 Ã— 2 = 140

70 Ã— 3 = 210

70 Ã— 4 = 280

70 Ã— 5 = 350

âˆ´ 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 itâ€™s an even number

(ii) 37

37 is not a multiple of 2

Hence itâ€™s an odd number

(iii) 50

50 is a multiple of 2

Hence itâ€™s an even number

(iv) 58

58 is a multiple of 2

Hence itâ€™s an even number

(v) 69

69 is not a multiple of 2

Hence itâ€™s an odd number

(vi) 144

144 is a multiple of 2

Hence itâ€™s an even number

(vii) 321

321 is not a multiple of 2

Hence itâ€™s an odd number

(viii) 253

253 is not a multiple of 2

Hence itâ€™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

âˆ´ 87 is not a prime number

(ii) 89

1 and 89 are the divisors of 89

âˆ´ 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

âˆ´ 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

âˆ´ 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 **

- Since every number has two factors i.e. 1 and itself .Hence there are no counting number having no factor at all.
- There is only one number which has exactly 1 factor i.e. is 1
- 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 oneâ€™s digits place in 2650

âˆ´ It is divisible by 2

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

âˆ´ It is not divisible by 2

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

âˆ´ It is divisible by 2

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

âˆ´ It is not divisible by 2

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

âˆ´ It is divisible by 2

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

âˆ´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

âˆ´ 733 is not divisible by 3

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

Since 12 is divisible by 3

âˆ´ 10038 is divisible by 3

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

Since 10 is not divisible by 3

âˆ´ 20701 is not divisible by 3

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

Since 27 is divisible by 3

âˆ´ 524781 is divisible by 3

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

Since 23 is not divisible by 3

âˆ´ 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

âˆ´ 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

- 618 is not divisible by 4 since the last two digits 18 is not divisible by 4
- 2314 is not divisible by 4 since last two digits is 14 is not divisible by 4
- 63712 is divisible by 4 since last two digits 12 is divisible by 4
- 35056 is divisible by 4 since last two digits 56 is divisible by 4
- 946126 is not divisible by 4 since last two digits 26 is not divisible by 4
- 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

âˆ´ 4965 is divisible by 5

(ii) Since 0 is in ones place

âˆ´ 23590 is divisible by 5

(iii) Since 8 is in ones place

âˆ´ 35208 is not divisible by 5

(iv) Since 5 is in ones place

âˆ´ 723405 is divisible by 5

(v) Since 4 is in ones place

âˆ´ 124684 is not divisible by 5

(vi) Since 0 is in ones place

âˆ´ 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

âˆ´ 2070 is divisible by 6

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

âˆ´ 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

âˆ´ 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

âˆ´ 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

âˆ´ 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

âˆ´ 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 â€“ 2 Ã— 6 = 70

70 is a multiple of 7

âˆ´ 826 is divisible by 7

(ii) 117

11 â€“ 2 Ã— 7 = -3

-3 is not multiple of 7

âˆ´ 117 is not divisible by 7

(iii) 2345

234 â€“ 2 Ã— 5 = 224

224 is a multiple of 7

âˆ´ 2345 is divisible by 7

(iv) 6021

602 â€“ 2 Ã— 1 = 600

600 is not multiple of 7

âˆ´ 6021 is not divisible by 7

(v) 14126

1412 â€“ 2 Ã— 6 = 1400

1400 is a multiple of 7

âˆ´ 14126 is divisible by 7

(vi) 25368

2536 â€“ 2 Ã— 8 = 2520

2520 is a multiple of 7

âˆ´ 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

âˆ´ 5790 is divisible by 10

(ii) 63215

Here the ones digit is 5

âˆ´ 63215 is not divisible by 10

(iii) 55555

Here the ones digit is 5

âˆ´ 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 â€“ 7 = 0

âˆ´ 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 â€“ 5 = 11

âˆ´ 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 â€“ 7 = 3

âˆ´ 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 â€“ 14 = 0

âˆ´ 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 â€“ 15 = -11

âˆ´ 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

âˆ´ 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:

âˆ´ 12 = 2 Ã— 2 Ã— 3

= 2^{2 }Ã— 3

The prime factorization of 12 = 2^{2 }Ã— 3

**2. 18**

** Solution**

** We have:**

âˆ´ 18 = 2 Ã— 3 Ã— 3

= 2 Ã— 3^{2}

The prime factorization of 18 = 2 Ã— 3^{2}

**3. 48**

** Solution**

We have:

âˆ´ 48 = 2 Ã— 2 Ã— 2 Ã— 2 Ã—3

= 2^{4} Ã— 3

The prime factorization of 48 = 2^{4} Ã— 3

**4. 56**

** Solution**

We have:

âˆ´ 56 = 2 Ã— 2 Ã— 2 Ã— 7

= 2^{3 }Ã— 7

The prime factorization of 56 = 2^{3 }Ã— 7

**5. 90**

** Solution**

We have:

^{ }

âˆ´ 90 = 2 Ã— 3 Ã— 3 Ã— 5

= 2 Ã— 3^{2 }Ã— 5

The prime factorization of 90 = 2 Ã— 3^{2 }Ã— 5

**6. 136**

** Solution**

We have:

âˆ´ 136 = 2 Ã— 2 Ã— 2 Ã— 17

= 2^{3 }Ã— 17

The prime factorization of 136 = 2^{3 }Ã— 17

**7. 252**

** Solution**

We have:

âˆ´ 252 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 7

= 2^{2 }Ã— 3^{2 }Ã— 7

The prime factorization of 252 = 2^{2 }Ã— 3^{2 }Ã— 7

** 8. 420**

** Solution**

We have:

âˆ´ 420 = 2 Ã— 2 Ã— 3 Ã— 7 Ã— 5

The prime factorization of 420 = 2^{2 }Ã— 3 Ã— 5 Ã— 7

**9.637**

** Solution**

We have:

âˆ´ 637 = 7 Ã— 7 Ã— 13

= 7^{2 }Ã— 13

The prime factorization of 637 = 7^{2 }Ã— 13

**10. 945**

** Solution**

We have:

âˆ´ 945 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 7

= 3^{3 }Ã— 5 Ã— 7

The prime factorization of 945 = 3^{3 }Ã— 5 Ã— 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**

** **

The prime factors of 84 = 2 Ã— 2 Ã— 3 Ã— 7 = 2^{2 }Ã— 3 Ã— 7

The prime factors of 98 = 2 Ã— 7 Ã— 7 = 2 Ã— 7^{2}

âˆ´ HCF = 2Ã— 7 = 14

**2. 170, 238**

** Solution**

** **

The prime factors of 170 = 2 Ã— 5 Ã— 17

The prime factors of 238 = 2 Ã— 7 Ã— 17

âˆ´ HCF = 2 Ã— 17 = 34

** 3. 504,980**

** Solution**

The prime factors of 504 = 2^{3 }Ã— 3^{2 }Ã— 7

The prime factors of 980 = 2^{2 }Ã— 5 Ã— 7^{2}

âˆ´ HCF = 2^{2 }Ã— 7 = 28

**72, 108, 180**

**Solution**

The prime factors of 72 = 2^{3 }Ã— 3^{2 }

The prime factors of 108 = 2^{3 }Ã—3

The prime factors of 180 = 2^{2 }Ã— 3^{2 }Ã—5

âˆ´ HCF = 2^{2 }Ã— 3^{2 }= 36

**84, 120, 138**

**Solution**

The prime factors of 84 = 2^{2 }Ã— 3 Ã—7

The prime factors of 120 = 2^{3 }Ã— 3 Ã— 5

The prime factors of 138 = 2 Ã— 3 Ã— 23

âˆ´ HCF = 2 Ã— 3 = 6

**106,159,371**

**Solution**

The prime factors of 106 = 2 Ã— 53

The prime factors of 159 = 3 Ã— 53

The prime factors of 371 = 7 Ã— 53

âˆ´ HCF = 53

**272, 425**

**Solution**

The prime factors of 272 = 2^{4} Ã— 17

The prime factors of 425 = 5^{2} Ã— 17

âˆ´ HCF = 17

**8**.**144, 252,630**

** Solution**

The prime factors of 144 = 2^{4 }Ã— 3^{2}

The prime factors of 252 = 2^{2 }Ã— 3^{2 }Ã— 7

The prime factors of 630 = 2 Ã— 3^{2 }Ã— 5

âˆ´ HCF = 18

**9. 1197, 5320, 4389**

** Solution**

The prime factors of 1197 = 3^{2 }Ã— 7 Ã— 19

The prime factors of 5320 = 2^{3 }Ã— 5 Ã— 7 Ã— 19

The prime factors of 4389 = 3Ã— 7 Ã— 11 Ã— 19

âˆ´ HCF = 133

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

** 10. 58, 70**

** Solution**

âˆ´ The common factor of 58, 70 =2

** 11. 399, 437**

** Solution**

âˆ´ The Common factor of 399, 437 = 19

** 12. 1045, 1520**

** Solution**

âˆ´ The HCF of 1045, 1520 = 95

** 13. 1965, 2096**

** Solution**

âˆ´ Common factor of 1965, 2096 = 131

** 14. 2241, 2324**

**Solution**

âˆ´ The Common factor of 2241, 2324 = 83

** 15. 658, 940, 1128**

**Solution**

âˆ´ Common factor of 658, 940, 1128 = 94

**754, 1508, 1972**

**Solution**

âˆ´ Common factor of 754, 1508, 1972 = 58

**17. 391, 425, 527**

**Solution**

âˆ´ Common factor of 391, 425, 527 = 17

**18. 1794, 2346, 4761**

**Solution**

âˆ´ Common factor of 1794, 2346, 4761 is 69

** Show that the following primes are co-primes:**

** 19. 59, 97**

** Solution**

59 = 59Ã— 1

97 = 97 Ã— 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

** 20. 161,192**

** Solution**

161 = 7 Ã— 23

192 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã—3

= 2^{6 }Ã—3 Ã—1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

** 21.343, 432**

** Solution**

343 = 7Ã— 7 Ã— 7

= 7^{3 }Ã— 1

432 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 3

= 2^{4 }Ã— 3^{3 }Ã— 1

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**22. 512, 945**

** Solution**

512 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2

= 2^{9}

945 = 3 Ã— 3 Ã— 3 Ã— 5 Ã— 7

= 3^{3 }Ã— 5 Ã— 7

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**23. 385, 621**

** Solution**

385 = 5 Ã— 7 Ã— 11 Ã— 1

621 = 3 Ã— 3 Ã— 3 Ã— 23 Ã— 1

= 3^{3 }Ã— 23

Here common factor = 1

HCF = 1

Hence the given numbers are co primes

**24. 847, 1014**

** Solution**

847 = 7 Ã— 11 Ã— 11 Ã— 1

= 7 Ã— 11^{2 }Ã— 1

1014 = 2 Ã— 3 Ã— 13 Ã— 13 Ã— 1

= 2 Ã— 3 Ã— 13^{2 }Ã— 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

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

## Exercise 2E PAGE NO: 40

**1. 42, 63**

**Solution**

42 = 2 Ã— 3 Ã— 7

63 = 3 Ã— 3Ã— 7

= 3^{2 }Ã— 7

âˆ´ LCM of 42 and 63 = 2 Ã— 3^{2 }Ã— 7

= 2 Ã— 9 Ã— 7

= 18 Ã— 7

= 126

**2. 60, 75**

**Solution**

60 = 2 c 2 Ã— 3 Ã— 5

= 2^{2 }Ã— 3 Ã— 5

75 = 3 Ã— 5 Ã— 5

= 3 Ã— 5^{2}

âˆ´ LCM of 60 and 75 = 2^{2 }Ã— 3 Ã— 5^{2}

= 4 Ã— 3 Ã— 25

= 12 Ã— 25

= 300

**3. 12, 18, 20**

**Solutions**

12 = 2 Ã— 2 Ã— 3 = 2^{2 }Ã— 3

18 = 2 Ã— 3 Ã— 3 = 2 Ã— 3^{2}

20 = 2 Ã— 2 Ã— 5 = 2^{2 }Ã— 5

âˆ´ LCM of 12, 18, and 20 = 2^{2 }Ã— 3^{2 }Ã— 5

= 4 Ã— 9 Ã— 5

= 180

**4. 36, 60, 72**

**Solutions**

36 = 2 Ã— 2 Ã— 3 Ã— 3 = 2 ^{2 }Ã— 3^{2}

60 = 2 Ã— 2 Ã— 3 Ã— 5 = 2^{2 }Ã— 3 Ã— 5

72 = 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 = 2^{3 }Ã— 3^{2}

âˆ´ LCM of 36, 60 and 72 = 2^{3 }Ã— 3^{2 }Ã— 5

= 8 Ã— 9 Ã— 5

= 72 Ã— 5

= 360

**5. 36, 40, 126**

**Solutions**

36 = 2 Ã— 2 Ã— 3 Ã— 3 = 2^{2 }Ã— 3^{2}

40 = 2 Ã— 2 Ã— 2 Ã— 5 = 2^{3} Ã— 5

126 = 2 Ã— 3^{2 }Ã— 7

âˆ´ LCM of 36, 40 and 126 = 2^{3} Ã— 3^{2 }Ã— 5 Ã— 7

= 8 Ã— 9 Ã— 5 Ã— 7

= 72 Ã— 35

= 2520

**6. 16, 28, 40, 77**

**Solution**

âˆ´ LCM of given numbers = 2 Ã— 2 Ã— 2 Ã— 7 Ã— 2 Ã— 5 Ã— 11

= 2^{4 }Ã— 5 Ã— 7 Ã— 11

= 16 Ã— 35 Ã— 11

= 6160

**7. 28, 36, 45, 60**

**Solutions**

âˆ´ LCM of given numbers = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 7

= 2^{2 }Ã— 3^{2 }Ã— 5 Ã— 7

= 4 Ã— 9 Ã— 5 Ã— 7

= 36 Ã— 35

= 1260

**8. 144, 180, 384**

**Solutions**

âˆ´ LCM of given numbers = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 Ã— 8

= 2^{4 }Ã— 3^{2} Ã— 5 Ã— 8

= 16 Ã— 9 Ã— 5 Ã— 8

= 16 Ã— 45 Ã— 8

= 5760

**9. 48, 64, 72, 96, 108**

**Solution**

âˆ´ LCM of given numbers = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 Ã— 2 Ã— 3

= 2^{6 }Ã— 3^{3}

= 64 Ã— 27

= 1728

**Find the HCF and LCM of **

**1. 117, 221**

**Solution**

To find the HCF of 117 and 221

âˆ´ HCF of 117and 221 = 13

Since, LCM = product of numbers/HCF

=117 Ã— 221/13

= 9 Ã— 221

= 1989

âˆ´ LCM = 1989 and HCF = 13

**2. 234,572**

Solution

To find the HCF of 234 and 572

âˆ´ HCF of 234 and 572 = 26

Since, LCM = Product of numbers/ HCF

= 234 Ã— 572/ 26

= 9 Ã— 572

= 5148

âˆ´ LCM = 5148 and HCF = 26

**3. 693, 1078**

**Solution**

To find the HCF of 693 and 1078

âˆ´ HCF of 693 and 1078 = 77

Since, LCM = Product of numbers/ HCF

= 693 Ã— 1078 / 77

= 9 Ã— 1078

= 9702

âˆ´ LCM = 9702 and HCF = 77

**4. 145,232**

**Solution**

To find the HCF of 693 and 1078

âˆ´ HCF of 145 and 232 = 29

Since, LCM = product of numbers / HCF

= 145 Ã— 232 / 29

= 5 Ã— 232

= 1160

âˆ´ LCM = 1160 and HCF = 29

**5. 861, 1353**

**Solution**

To find HCF of 861 and 1353

âˆ´ HCF of 861 and 1353 = 123

Since, LCM = product of numbers /HCF

= 861 Ã— 1353 / 123

= 7 Ã— 1353

= 9471

âˆ´ LCM = 9471 and HCF = 123

**6. 2923, 3239**

**Solution**

To find HCF of 2923 and 3239

âˆ´ HCF of 2923 and 3239 = 79

Since, LCM = product of numbers/ HCF

= 2923 Ã— 3239 / 79

= 37 Ã— 3239

= 119843

âˆ´ 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 Ã· 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 oneâ€™s digits is divisible by 8 i.e. 176

176 Ã·Â 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) â€“ (2 + 2 + 2 + 2) = 8 â€“ 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 Ã— 19

Hence 323 is not a prime number

b) 361 can be written as 19 Ã— 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 donâ€™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 Ã— 2 Ã— 2 Ã— 2 Ã— 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

Here 144 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3

= 2^{4 }Ã— 3^{2}

198 = 2 Ã— 3 Ã— 3 Ã— 11

= 2 Ã— 3^{2 }Ã— 11

18 = 2 Ã— 3^{2} 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 2^{2 }Ã— 3 = 12

The factorization of 144, 180 and 192

Hence 144 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 Ã— 3 = 2^{4 }Ã— 3^{2}

180 = 2 Ã— 2 Ã— 3 Ã— 3 Ã— 5 = 2^{2 }Ã— 3^{2 }Ã— 5

192 = 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 2 Ã— 3 = 2^{6 }Ã— 3

Hence 2^{2 }Ã— 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

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 Ã— 5 Ã— 11

132 = 2 Ã— 2 Ã— 3 Ã— 11

Their common factors are 11 and 3 = 11 Ã— 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

### 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.