Table of Contents

**Highest Common Factors **

Highest common factor is the greatest number which will divide exactly into two or more numbers. For example 4 is the **highest common factor (HCF)** of 20 & 24.

i.e. 20 = 1, 2, (4), 5, 10, 20

24= 1, 2, 3, (4), 6, 8, 12, 24

**Example 1:**

Find the H.C.F of 24 & 78

__ __

__Method 1 __

Express each number as a product of its prime factors

Workings

2 24 2 78

2 12 2 36

2 6 2 18

3 3 3 9

3 3

24=2^{3}x3

78=(2^{3 }x 3) x 3

The H.C.F. is the product of the common prime factors.

HCF=2^{3}x3

=8×3=24

__ __

__Method II__

24=2x2x2x3

78=2x2x2x3x3

Common factor=2x2x2x3

HCF=24

**LCM: Lowest Common Multiple**

Multiples of 2 are =2,4,6,8,10,12,14,16,18,20,22,24

Multiples of 5 are 5,10,15,20,25,30,35,40

Notice that 10 is the lowest number which is a multiple of 2 & 5.10 is the lowest common multiple of 2& 5

Find the LCM of 20, 32, and 40

** **

**Method 1**

Express each number as a product of its prime factors

20=2^{2}x5

32=2^{5}

40=2^{2}x2x5

The prime factors of 20, 32 and 40 are 2 & 5 .The highest power of each prime factor must be in the LCM

These are2^{5} and 5

Thus LCM =2^{5} x5

=160

** **

**Method II **

2 20 32 40

2 10 16 20

2 5 8 10

4 5 4 5

5 5 1 5

1 1 1

LCM =2 x 2 x 2 x 4 x 5 = 160

** **

**Class work**

Find the HCF of:

(1) 28 and 42

(2) 504 and 588

(3) Find the LCM of 84 & 210

** **

## PERFECT SQUARES

A perfect square is a whole number whose square root is also a whole number .It is always possible to express a perfect square in factors with even indices.

9 = 3×3

25= 5×5

225 = 15×15

= 3x5x3x5

= 3^{2} x 5^{2}

9216 =96 ^{2}

=3^{2 }x 32 ^{2}

=3^{2} x 4^{2} X 8^{2 }

=3^{2}x2^{4} x2^{6 }

=3^{2} x2 ^{10}

** **

**Workings **

2 9216

2 4608

2 2304

2 1152

2 576

2 288

2 144

2 72

2 36

2 18

3 9

3 3

9216= 3^{2}x2^{10}

Example

Find the smallest number by which the following must be multiplied so that their products are perfect square

- 540
- 252

**Solution **

2 540

2 270

3 135

3 45

3 15 54=2^{2 }x 3^{3}x 5

5 5

1

The index of 2 even. The index of 3 and 5 are odd .One more 3 and one more 5 will make all the indices even. The product will then be a perfect square .The number required is 3×5 =__15__

- 2 252

2 126

3 63

3 21

7 7

1

252= 2^{2}x3^{2}x7

Index of 7 is odd, one more 7 will make it even.

Indices i.e. 2^{2}x 3^{2}x 7^{2}

**Therefore 7 is the smallest numbers required**

** **

**WEEKEND ASSIGNMENT**

- The lowest common multiple of 4, 6 and 8 is (a) 24 (b) 48 (c) 12 (d) 40
- Find the smallest number by which 72 must be multiplied so that its products will give a perfect square (a) 3 (b) 2 (c) 1 (d) 5
- The lowest common multiple of 4, 6 and 8 is (a) 24 (b) 48 (c) 12 (d) 40
- The H.C.F. of 8, 24 and 36 is ___ (a) 6 (b) 4 (c) 18 (d) 20
- The L.C.M. of 12, 16 and 24 is ___ (a) 96 (b) 48 (c) 108 (d) 24

**THEORY **

- Find the smallest number by which 162 must be multiplied so that its product will give a perfect square.
- Find the HCF and L.C.M. of the following figures

30 & 42

64 & 210

**See also**

WHOLE NUMBERS AND DECIMAL NUMBERS

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