GRADE : VI

SUBJECT: Maths

READING SECTION: Chapter:2 Properties of whole numbers

Multiples of a number

7 2

7 × 𝟏 = 7 2 × 1 = 2

7 × 2 = 14 2 × 2 = 4

7 × 3 = 21 2 × 3 = 6

7, 14, 21, 28…….. 2, 4, 6 ………. are multiple of 2

are the multiples of 7

Even

numbers

Points to Remember:

1) 1 is a factor of every number

1 x 6 = 6

1 x any number = number itself

any number ÷ 1 = number itself

8 ÷ 1 = 8

2) Every number is a factor of itself.

6 x 1 = 6

any number x 1 = number itself

any number ÷ itself = 1

8 ÷ 8 = 1

3) The factor of a number is always less than or

equal to the number.

Factor of 12: 1, 2, 3, 4, 6 and 12

1 < 12

2 < 12

3 < 12

4 < 12

6 < 12

12= 12

Factors of 10, 12 and 32 Multiples of 10, 12 and 32

Here,

10 = 1, 2, 5, 10 are factors of 10.

12 = 1, 2, 3, 4, 6, 12 are factors of 12

32 = 1, 2, 4, 8, 16, 32are factors of 32

Here,

Multiples of 10 ; 10, 20, 30, 40, 50…….are

multiples of 10.

Multiples of 12; 12, 24, 36, 48, 60 ……. are

the multiple of the 12

Multiples of 32: 32, 64, 96, 128, 160……. are

the multiple of 32

Factors and multiples:

Prime factorization for the given numbers.

36 45 70

36 = 2×2×3×3 45=3×3×5 70= 2×5×7

Note: To get prime factors ,we divide by only prime numbers like 2,3,5 ,7 and so on

1. Write all the possible factors of these numbers.

a) 16 b) 35 c) 64 d) 18 e) 140

2. Write the first five multiples of these numbers.

a) 7 b) 9 c) 11 d) 15 e) 25

3. Factorize the following numbers using prime factorization process.

a) 18 b) 24 c) 72 d) 100 e) 150

Exercise:1

Worked Out Example:

HCF = Product of their common prime

factors HCF stands for Highest Common Factor

Three methods of finding HCF

Writing all the

possible factors Prime factorization

method Division method

Writing all the possible factors

a) 16, 20

Factors of 16= 1, 2, 4, 8, 16

Factors of 20= 1, 2, 4, 5, 10, 20

Here, common factors= 1, 2, 4

(among these common factors, 4 is

the highest )

Therefore H.C.F of 16,20=4

b) 14, 28,42

Factors of 14= 1, 2, 4, 7, 14

Factors of 28= 1, 2, 4, 7, 14, 28

Factors of 42= 1, 2, 3, 6, 7, 14, 21,

42

Here, common factors = 1,2,7,14

(among these common factors, 14 is

the highest )

Therefore H.C.F of 14,28 42= 14

Method One

Prime factorisation method

a)16,20 b) 14, 28, 42

H.C.F= 2 × 2 H.C.F= 2× 7

= 4 = 14

Method Two

Method Three

Divide the bigger number by the smaller one, divide each

divisor by corresponding remainder till the remainder comes to

zero

a) 16, 20

Note: The last divisor is the H.C.F.

HCF = 4

Step 1: Divide the bigger number by smaller one, divide each

divisor by corresponding remainder till remainder comes zero.

Here HCF

of 60 and

96 = 12

Step 2: Again, divide the third number by the obtained HCF and

follow the previous step till the remainder comes zero.

Hence HCF of 60, 96 and 150 = 6.

1.Write all the possible factors of the following pairs of numbers , circle the common factors and

select the H.C.F

a) 8 , 12 b) 16, 24 c) 20, 30 d) 24 , 36 e) 15, 30, 45

2. Find the H.C.F of the following pairs of numbers by prime factorization method.

a) 12, 16 b) 15, 20 c) 24, 30 d) 44, 66

e) 12, 18, 24 f) 27, 45, 63 g) 90, 150, 180

3. Find the H.C.F of the following pairs of numbers by division method.

a) 16, 24 b) 15, 20 c) 26, 39 d) 54, 90

e) 80, 120 f) 27, 45, 63 g) 80, 120, 200

Exercise: 2

LCM stands for Lowest Common Multiple

LCM = Product of their common prime factors and

remaining factors

Three methods of finding LCM

Writing the first

ten multiples Division

process. Prime factorization

method

Let’s find L.C.M of : a) 8, 12 b) 6, 10, 15

Writing the first ten multiples

a) 8, 12

Multiples of 8= 8, 16, 24, 32, 40, 48,

56, 64, 72, 80

Multiples of 12= 12, 24, 36, 48, 60, 72,

84, 96 ,108,

Here, common multiples are 24, 48 and

72

(among these common multiples 24 is

the lowest)

Therefore L.C.M of 8, 12= 24

b) 6, 10, 15

Multiples of 6= 6, 12, 18, 24, 30,

36, 42, 48, 54, 60

Multiples of 10= 10, 20, 30, 40, 50,

60, 70, 80, 90, 100 , 120

Multiples of 15= 15, 30, 45, 60, 75,

90, 105, 120, 135, 150

Here common multiples are 30, 60

(among these common multiples 30

is the lowest)

Therefore L.C.M of 6, 10, 15 = 30

Method One

Method Two

Division Process

a) 8, 12

L.C.M = 2 × 2 × 2 × 3

= 24

b) 6, 10, 15

L.C.M = 2 × 3 × 5

= 30

Prime factorization method

a) 8, 12

Method Three

8 = 2 x 2 x 2

12= 2 x 2 x 3

L.C.M = 2 x 2 x 2 x 3

= 24

Take 2 x 2 from first

and second number

then, multiply the

remaining number first

and remaining number

of second

b) 24, 36, 60

24 = 2 x 2 x 2 x 3

36 = 2 x 2 x 3 x 3

60 = 2 x 2 x 3 x 5

L.C.M = 2 x 2 x 2 x 3 x 3 x 5

= 360

here, among 2x2x2 and 2x2, select

the highest value i.e 2x2x2.

Similarly among 3 and 3x3 select

3x3

5 is only lef so take it, because there

is no option for 5.

1.Write the first ten multiples of the following pairs of numbers and select the L.C.M

a) 2, 4 b) 6, 8 c) 7, 14 d) 10 , 20 e) 4, 6,

2. Find the L.C.M of the following pairs of numbers by division method.

a) 8, 10 b) 9, 15 c) 16, 24 d) 22, 33

e) 8, 12, 16 f) 18, 24, 36 g) 40, 45, 60

3. Find the L.C.M of the following pairs of numbers by prime factorisation method.

a) 12, 15 b) 30, 45 c) 32, 48 d) 16, 24, 32

e) 25, 50, 75 f) 36, 48, 60 g) 120, 144, 160

Exercise: 3

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