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**Brilliant’s Composite Mathematics Class 6 Solutions Chapter 4 Factors and Multiples**

Welcome to NCTB Solutions. Here with this post we are going to help 6th class students for the Solutions of Brilliant’s Composite Mathematics Class 6 Math Book, Chapter 4, Factors and Multiples. Here students can easily find step by step solutions of all the problems for Factors and Multiples, Exercise 4.1, 4.2, 4.3, 4.4 and 4.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.

**Factors and Multiples Exercise 4.1 Solution**

**Question no – (1) **

**Solution :**

**(i)** All the possible factors of 36

= 1 × 36 = 2 × 18 = 3 × 12 = 4 × 9 = 6 × 6

**∴** Factor are = 1, 2, 3, 4, 6, 9, 12, 18 26

**(ii)** All the possible factors of 25

= 25 × 1 = 5 × 5

**∴** Factors are = 1, 5, 25

**(iii)** All the possible factors of 50

= 50 = 1 × 50 = 2 × 25 = 5 × 10

**∴** Factors are = 1, 2, 5, 10, 25, 50

**(iv)** All the possible factors of 60

= 1 × 60 = 2 × 30 = 3 × 20 = 4 × 15 = 5 × 12 = 6 × 10

**∴** Factors are = 1, 2, 3, 4, 5, 6, 10, 15, 20, 30, 60

**Question no – (2) **

**Solution :**

**(i)** The first five multiples of 14 are 14, 28, 42, 56, 70.

**(ii)** The first five multiples of 22 are 22, 66, 44, 88, 110

**(iii)** The first five multiples of 31 are 31, 62, 93, 124, 155

**(iv)** The first five multiples of 38 are 38, 76, 114, 152, 190

**(v)** The first five multiples of 45 are 45, 90, 135, 225

**(vi)** The first five multiples of 62 are 62, 124, 186, 248, 310

**Question no – (3) **

**Solution :**

**(i) 1625**

= 25 is a factor of 1625

**(ii) 18220**

= 25 is not a factor of 18220

**(iii) 167225**

= 25 is a factor of 167225

**(iv) 186550**

= 25 is a factor of 186550

**Question no – (4) **

**Solution :**

**(i)** All prime numbers between 1 and 40 are 2, 3, 5, 7, 11, 13, 17, 19, 123, 29, 31, 37

**(ii)** All prime numbers between 50 and 100 are 53, 59, 61, 67, 71, 73, 79, 83, 89, 97

**(iii)** All prime numbers between 87 and 100 are 89 and 97

**(iv)** All prime numbers between 55 and 75 are 59, 61, 67, 71, 73

**Question no – (5)**

**Solution :**

One even number is prime 2

**Question no – (6)**

**Solution :**

Prime Number :

A number which have exactly two factors is called a prime number.

Examples : 1, 2, 3, 5, 7 are prime numbers.

**Question no – (7)**

**Solution :**

Prime numbers are – (ii) 23 (iii) 31 (iv) 43

Composite. umber are – (i) 21 (v) 87 (vi) 105 (vii) 187

**Question no – (8)**

**Solution :**

Composite number :

A number which have more than two factors is called a composite number.

Yes, a composite number be odd.

Smallest odd composite number is 9

**Question no – (9)**

**Solution :**

The smallest even prime number is 2

**Question no – (10)**

**Solution :**

Two numbers are and said to be co-prime numbers if they do not have common factor other then 1.

Examples : (2, 3) (3, 4), (4, 5) (5, 6)

The pair of numbers 9, 10 are co-prime numbers while 9 is not a prime number.

**Question no – (11)**

**Solution :**

Twin Primes :

Two prime number which by 2 are called twin primes.

**Question no – (12) **

**Solution :**

**(i)** 40 as a sum of two odd primes,

= 3 + 37

**(ii)** 80 as a sum of two odd primes,

= 7 + 73

**(iii)** 100 as a sum of two odd primes,

= 100 = 3 + 97

**(iv)** 96 as a sum of two odd primes,

= 7 + 89

**(v)** 56 as a sum of two odd primes,

= 3 + 53

**Question no – (13) **

**Solution :**

**(i)** 35 as sum of three odd prime numbers,

= 23 + 7 + 5

**(ii)** 49 as sum of three odd prime numbers,

= 3 + 5 + 41

**(iii)** 31 as sum of three odd prime numbers,

= 19 + 5 + 7

**(iv)** 63 as sum of three odd prime numbers,

= 7 + 13 + 43

**Question no – (14)**

**Solution :**

list of five consecutive numbers, none of which is prime – 61, 62, 63, 64, 65,

– 61 is a prime number.

**Question no – (15)**

**Solution :**

**(i)** No, natural number having no factor at all.

**(ii)** Only One (1) natural numbers are there having exactly one factor.

**(iii)** The natural numbers which are having exactly one factor are = 4, 9, 25, 49

**Question no – (16) **

**Solution :**

**(i)** There is no natural number having no factor at all.

= Given statement is **False**

**(ii)** The sum of two prime numbers is not always a prime number.

= Given statement is **True**

**(iii)** If two numbers are co-primes, at least one of them must be a prime.

= Given statement is **False**

**(iv)** 6 is a perfect number.

= Given statement is **True**

**(v)** Numbers having more than two factors are known as composite numbers.

= Given statement is **True**

**(vi)** 1 is neither a prime nor a composite number.

= Given statement is **True**

**(vii)** 2 is not a the lowest prime number.

= Given statement is **False**

**(viii)** The only even prime number is 2.

= Given statement is **True**

**Factors and Multiples Exercise 4.2 Solution**

**Question no – (1) **

**Solution :**

As we know that,

If the last digit of the number is 0, 2, 4, 6, and 8 then the number is divisible by 2

A number is divisible by 4 if the last two digit of the number is divisible by 4 or the whole number is divisible by 4

A number is divisible by 8 if its last three digits are divisible by 8 then the number is divisible by 8

Hence,

(i), (ii) (iii), (iv) (iv), (iv) (vii) (viii) (ix), (x) all are divisible by 2

(iii), (iv) (v), (vi), (ix) are divisible by 4

(iv), (ix) divisible by 8

**Question no – (2) **

**Solution :**

As we know that,

A number is divisible by 3 if the sum of digits is divisible by 3

A number is divisible by 6 if the number is divisible by both the number 2 and 3

A number is divisible by 9 if the sum of all digits is divisible by 9

Hence,

(i), (ii) (iii), (iv), (v), (vi) (vii) (viii), are divisible by 3

(i), (ii) (iii), (v), (vi) are divisible by 6

(i), (ii) (iv) (v) are divisible 9

**Question no – (3) **

**Solution :**

As we know that,

A number is divisible by 5 if the last digit of the given number is 0 or 5

A number is divisible by 10 if its unit digit place is ‘0’

Therefore,

(i), (ii), (iii), (iv), (v), (vi) are divisible by 5

(i), (iii), (vi) are divisible by 10

**Question no – (4) **

**Solution :**

As we know that,

A number is divisible by 11 if the sum of odd place and the sum of even place. And then their difference is divisible by 11.

Therefore,

(i), (ii), (iii) (iv), (v) are divisible by 11.

**Factors and Multiples Exercise 4.3 Solution : **

**Question no – (1) **

**Solution :**

**(i) 36**

Now,

**∴** 36 = 2 × 2 × 3 × 3

**(ii) 48**

Now,

**∴** 48 = 2 × 2 × 2 × 2 × 3

**(iii) 60**

Now,

**∴ **60 = 2 × 2 × 3 × 5

**(iv) 84**

Now,

**∴** 84 = 2 × 2 × 3 × 7

**(v) 132**

Now,

**∴** 132 = 2 × 2 × 3 × 11

**(vi) 225**

Now,

**∴** Prime factorisation of 225 = 5 × 5 × 3 × 3

**(vii) 441**

Now,

∴ Prime factorisation of 441 = 3 × 3 × 7 × 7

**(viii) 525**

Now,

**∴** Prime factorisation of 525 = 5 × 5 × 3 × 7

**(ix) 900**

Now,

**∴** Prime factorisation of 900 = 3 × 2 × 2 × 2 × 5 × 5

**(x) 1350**

Now,

**∴** Prime factorisation of 1350 = 3 × 3 × 3 × 2 × 5 × 5

**Question no – (2)**

**Solution :**

As we know,

Largest 4 digit number is 9999

Now,

**∴** 9999 = 3 × 3 × 11 × 101

**Question no – (3)**

**Solution :**

As we know,

The Smallest 4 digit number is 1000

Now,

**∴** 1000 = 2 × 2 × 2 × 5 × 5 × 5

**Question no – (4)**

**Solution :**

7 ; 13, 19

The difference between any two consecutive prime number is 6.

**Question no – (5)**

**Solution :**

1 and the number itself are not included in the prime factorisation of a composite number.

**Factors and Multiples Exercise 4.4 Solution : **

**Question no – (1) **

**Solution :**

**(i) 12 and 16**

**∴** 12 = 2 × 2 × 3

**∴** 16 = 2 × 2 × 2 × 2

Therefore, H.C.F = 2 × 2 = 4

Hence, the HCF of 12 and 16 is 4

**(ii) 36 and 60**

**∴** 36 = 4 × 9 = 2 × 2 × 3 × 3

**∴** 60 = 2 × 2 × 3 × 5

**∴** H.C.F = 2 × 2 × 3 = 12

Hence, the HCF of 36 and 60 is 12

**(iii) 16 and 24**

**∴ **16 = 2 × 2 × 2 × 2

**∴** 24 = 2 × 2 × 2 × 3

**∴** H.C.F = 2 × 2 × 2 = 8

Hence, the HCF of 16 and 24 is 8

**(iv) 47 and 61**

**∴** 47 = 47 × 1

**∴** 61 = 61 × 1

**∴** H.C.F = 1

Thus, the HCF of 47 and 61 is 1

**(v) 40 and 48**

**∴** 40 = 2 × 2 × 2 × 5

**∴** 48 = 2 × 2 × 2 × 2 × 3

**∴** H.C.F = 2 × 2 × 2 = 8

Thus, the HCF of 40 and 48 is 8

**(viii) 60, 90 and 120**

**∴** 60 = 2 × 2 × 3 × 5

**∴** 90 = 3 × 3 × 0 × 2 × 5

**∴** 120 = 2 × 2 × 2 × 3 × 5

**∴** H.C.F = 2 × 3 × 5 = 30

Hence, the HCF of 60, 90 and 120 is 30

**(ix) 106, 159, 26**

Now,

**∴** 106 = 2 × 53

**∴** 159 = 3 × 53

**∴** 265 = 5 × 53

**∴** H.C.F = 53

Thus, the HCF of 106, 159, 26 is 53

**(x)** Given numbers, 101, 573, 1079

Now,

= 101 = 1 × 10

= 573 = 3 × 191 × 1

= 1079 = 1 ×1079

Hence, the HCF of 101, 573, 1079 is 1

**Question no – (3)**

**Solution :**

Given numbers are = 84 and 90.

Now,

**∴** The greatest number will be 6

Thus, the largest number will be 138

**Question no – (4)**

**Solution :**

H.C.F of any two consecutive number is 1

**Question no – (5)**

**Solution :**

**(i)** Given, 576/252

= 16/7

**∴** [H.C.F = 36]

**(ii)** Given, 522/1276

= 9/22

[H.C.F = 58]

**(iii)** Given, 1379/357

= 197/57

**∴** [H.C.F = 7]

**(iv)** Given, 4130/7021

= 590/1003

**∴** [H.C.F = 7]

**Question no – (7)**

**Solution :**

In the question,

Two milk containers contain = 300 litres and 450 litres

Now,

= 300 = 3 × 2 × 2 × 5 × 5

= 450 = 3 × 3 × 5 × 2 × 5

**∴** Maximum capacity

= 2 × 5 × 5 × 3

= 150 litres

Hence, the maximum capacity of container will be 150 litres

**Question no – (8)**

**Solution :**

We should find H.C.F of,

= (152 – 2) = 150,

= (144 – 4) = 140,

= (211 – 1) = 210

Now,

**∴** The largest number will be 14

**Question no – (9)**

**Solution :**

According to the question,

= 9 m 75 cm = 975 cm

= 8 m 25 cm = 825 cm

= 5 m = 25 cm = 252 cm

Now,

**∴** The largest tape which can measure the these dimensions of the room exactly = 75 cm.

**Question no – (10)**

**Solution :**

We find the H.C.F of,

= 245 – 5 = 240 ,

= 1029 – 5 = 1025

Now,

**∴** HCF of 240, 1024 is 16

**Question no – (11)**

**Solution :**

First we find the HCF of –

= (398 – 7) = 391,

= (542 – 15) = 527,

= (436 + 1) = 425

Therefore,

**∴** H.C.F will be = 17

**Question no – (12)**

**Solution :**

First we find the HCF of,

= (533 – 13) = 520,

= (1445 – 15) = 1430

Therefore,

**∴** H.C.F = 130

**Question no – (13)**

**Solution :**

First we find the H.C.F of ,

= (365 – 8) = 357

= (512 – 8) = 540

= (323 – 8) = 35

Now,

Therefore, the H.C.F will be = 21

**Factors and Multiples Exercise 4.5 Solution : **

**Question no – (1) **

**Solution :**

**(iii) 40, 36, 126**

Now,

**∴** LCM = 2 ×4 × 5 × 7 × 9

= 2520

So, the LCM of 40, 36 and 126 is 2520.

**(vi) 108, 135, 162**

Now,

= LCM = 2 × 2 × 3 × 3 × 5 × 9

= 1620

Thus, the LCM of 108, 135, and 162 is 1620

**(vii) 112, 168, 266**

Now,

= LCM = 2 × 2 ×3 × 4 × 7 × 19

= 6384

Thus, the LCM of 112, 168, 266 is 6384

**(viii) 180, 384, 144**

Now, LCM,

= LCM = 4 × 4 × 5 × 9 × 24 = 5760

Thus, the LCM of 180, 384 and 144 is 5760

**(ix) 24, 36, 60, 84, 135**

Now the LCM,

**∴** LCM = 2 × 3 × 3 × 3 × 4 × 5 × 7 = 7560

Thus, the LCM of 24, 36, 60, 84 and 135 is 7560

**Question no – (2)**

**Solution :**

**(i) 35, 40;**

= 35 = 5 × 7

40 = 5 × 2 × 2 × 2

HCF = 5

LCM = 5 × 2 × 2 × 2 × 7 = 280

Product of numbers = 35 × 40

= 1400

LCM × HCF = 280 × 5 = 1400

**(ii) 24, 40;**

= 24 = 4 × 2 × 3

40 = 4 × 5 × 2

HCF = 4 × 2 = 8

LCM = 4 × 2 × 3 × 5 = 120

24 × 40 = 960

120 × 8 = 960

**(iii) 50, 60;**

= 50 = 5 × 5 × 2

60 = 6 × 5 × 2

HCF = 2 × 50 = 10

LCM = 5 × 2 × 5 × 6 = 300

50 × 60 = 3000

and 300 × 10 = 3000

**(iv) 45, 105;**

= 45 = 5 × 3 × 3

105 = 5 × 3 × 7

HCF = 5 × 3 = 15

LCM = 5 × 3 × 3 × 7 = 315

45 × 105 = 4725

1315 × 15 = 4725

**(v) 117, 221.**

= 117 = 3 × 3 × 13

221 = 13 × 17

LCM = 3 × 3 × 13 × 17 = 1989

117 × 221 = 25857

and 13 × 1989 = 25857

**Question no – (3) **

**Solution :**

**(i) 4/55, 9/110, 13/165**

Now the LCM,

**∴** LCM = 11 × 5 × 2 × 3 = 330

= 4 × 6/55 × 6, 9 × 3/110 × 3, 13 × 2/165 × 2

= 24/330, 27/330, 26/330

**(ii) 4/15, 7/30, 8/45**

Now the LCM,

**∴** LCM = 15 × 2 × 3 = 90

= 4 × 6/15 × 6, 7 × 3/30 × 3, 8 × 2/45 × 2

= 24/90, 21/90, 16/90

**Question no – (4)**

**Solution :**

In the given question,

HCF and LCM of two numbers are 25 and 450

One of the numbers is = 75

Other number = ?

**∴** Other number = LCM × HCF/one number

= 450 × 25/75

= 150

Thus, the other number will be 150

**Question no – (5)**

**Solution :**

According to the question,

HCF of two numbers is = 24

Product is = 7200,

LCM = ?

**∴** LCM = product of two number/HCF

= 7200/24

= 300

Hence, their LCM will be 300

**Question no – (6)**

**Solution :**

Given in the question,

LCM of two numbers is 720

Product of the numbers is 2160

HCF = ?

**∴** HCF = Product of two number/LCM

= 2160/720

= 3

Hence, their HCF will be 3

**Question no – (7)**

**Solution :**

In the given question,

Smallest number which when divided by = 25, 40 and 60

leaves remainder = 7 in each case

Now, the LCM,

**∴** LCM = 5 × 4 × 5 × 2 × 3 = 600

**∴** Required number,

= 600 + 7

= 607

Hence, the smallest number will be 607.

**Question no – (8)**

**Solution :**

According to the question,

D to A = 40

A to D = 45 min

Now,

∴ Time take = 5 × 8 × 9 in

= 360 min = 6 hr

**∴** (10 : 15 + 6) = 4 : 15 p.m. two buses pass the same bridge.

**Question no – (9)**

**Solution :**

= 204/14

= 102/7

**∴** HCF of any number must divided their LCM. Here 14 does not divided 204.

**Question no – (10)**

**Solution :**

In the question we get,

A boy saves daily = Rs 4.65

∴ 4.65 = 465/100

in 2 days = 4.65 × 2 = 9.30

in 20 days = 4.65 × 20/100 = 93

Thus, in 20 days he will able to save an exact number of rupees.

**Question no – (11)**

**Solution :**

**∴** LCM = 3 × 3 × 4 × 5 = 180 min

**∴** 180 min = 3 hour after they next toll together.

Thus, they will next toll together in 3 hour.

**Question no – (12)**

**Solution :**

73 = 73 × 1

79 = 79 × 1

**∴** The HCF = 1

**∴** The LCM = 73 × 79 = 9761

Thus, their HCF is 1 and LCM is 9761

**Question no – (13)**

**Solution :**

In the question we get,

The HCF and LCM of two numbers are 19 and 114.

One of the number is = 38,

Other number = ?

**∴** Other number = LCM × HCF/one number

= 19 × 114/38

= 57

Hence, the other number will be 57

**Question no – (14)**

**Solution :**

In the question we get,

There are four bells which toll at intervals = 3, 7, 12 and 14 second

Now the LCM,

**∴** LCM = 2 × 2 × 3 × 7 = 84 = 1 min 24 sec

**∴** The time they will strike together next,

= 12 : 01 : 24 in 14 min

= 14 × 60/84

= 10 times

So, they will next toll together in 1 min 24 sec and 10 times.

**Question no – (15)**

**Solution :**

First, LCM of 40 and 30,

**∴** LCM = 5 × 2 × 4 × 3

= 120 sec

= 2 min

Hence, they will change in 2 minutes.

**Question no – (16)**

**Solution :**

First the LCM,

**∴** LCM = 5 × 2 × 8 × 9 × 17 = 12240

= 122 m 40 on

**∴** At 122 m 40 cm from the starting point they again step off together.

**Question no – (17)**

**Solution :**

First, the LCM of 220 and 300,

**∴** LCM = 10× 2 × 11 × 15 = 3300

Thus, 3300 metre far from the road.

**Question no – (18)**

**Solution :**

First, we find LCM of 27, 35, 25 and 21

**∴** L.C.M = 3 × 5 × 5 × 7 × 9 = 4725

**∴** The smallest number,

= 4725 – 3

= 4722

**Question no – (19)**

**Solution :**

First we find the LCM of = 48, 60 and 64.

**∴** L.C.M = 2 × 2 × 2 × 2 × 3 × 4 × 5

= 960

Hence, the required greatest number will be 960.

**Question no – (20)**

**Solution :**

From the question we get,

= 7m 80 cm = 780 cm

= 6 m 60 cm = 660

Now, the LCM of 780 and 660

**∴ **LCM = 6 × 10 × 13 × 11 = 8580

Hence, Number of such tile will be 8580

**Question no – (21)**

**Solution :**

From the question we get,

= 6 m 50 cm = 650 cm

= 3 m 50 cm = 350 cm

= 3 m 50 cm = 300 cm

Now, The LCM of 650, 350, 300,

∴ L.C.M = 10 × 13 × 7 × 6 × 5 = 27300

Hence, the largest tape = 27300 cm = 273 m

**Question no – (22)**

**Solution :**

First we find the LCM of 2, 3, 4, 5, 6 and 7

**∴** L.C.M = 2 × 2 × 3 × 5 × 7 = 920

We know, Greatest 4 digit number = 9999

**∴** Required number,

= 9999 – 339

= 9660

Thus, the greatest number will be 9660

**Question no – (23)**

**Solution :**

L.C.M. of 777, 819 and 4329

**∴** L.C.M = 3 × 3 × 7 × 13 × 37 = 30303

Thus, the LCM of 777, 819 and 4329 is 30303

**Question no – (24)**

**Solution :**

GCM + LCM = 375

GCM + 24 × GCM = 375

= 25 GCM = 375

**∴** GCM = 375/25 = 15

**∴** LCM = 15 × 24 = 3160

**∴** Second number,

= LCM × GCM/First number

= 360 × 15/45

= 120

Hence, the other number will be 120

**Question no – (25)**

**Solution :**

Here we have to find the least square number,

Which is exactly divisible by 4 times 5, 6, 7, 8, 9, 10

= 20, 24, 28, 32, 36, 40

Now the LCM of 20, 24, 28, 32, 36, 40

**∴** L.C.M = 2 × 4 × 4 × 3 × 3 5 × 7

= 22 × 22 × 32 × 70

= 10080

**∴** The least square number,

= 10080 × 70

= 705600

**Question no – (26)**

**Solution :**

First LCM of 2, 3, 4, 5 and 6

**∴** L.C.M = 2 × 3 2 × 5 = 60

**∴** Least number of 4 digit = 1000

**∴** The least number of 4 digit divisible,

= 100 + (60 – 40)

= 100 + 20

= 1020

**∴** Required number,

= 1020 + 1

= 1021

**Question no – (27)**

**Solution :**

First we need to find the LCM of 12, 15, 20 or 54

**∴** L.C.M = 2 × 2 × 3 × 5 × 9 = 540

**∴** Required number,

= 540 + 4

= 544

Thus, the required number will be 544

**Next Chapter Solution : **