H.C.F & L.C.M AND PERFECT SQUARES

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=2³x3

78=(2³x 3) x 3

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

HCF=2³x3

=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=2²x5

32=2⁵

40=2²x2x5

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

These are2⁵ and 5

Thus LCM =2⁵ 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

= 3² x 5²

9216 =96 ²

=3²x 32 ²

=3² x 4² X 8²

=3²x2⁴ x2⁶

=3² 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= 3²x2¹⁰

Example

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

Solution

2 540

2 270

3 135

3 45

3 15 54=2²x 3³x 5

5 5

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

252= 2²x3²x7

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

Indices i.e. 2²x 3²x 7²

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

BASIC OPERATION OF INTEGERS

WEIGHT

VOLUME OF CYLINDER

AREA OF RIGHT ANGLED TRIANGLE