# Lesson Notes By Weeks and Term - Junior Secondary School 2

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

SUBJECT: MATHEMATICS

CLASS:  JSS 2

DATE:

TERM: 1st TERM

WEEK THREE

TOPIC: H.C.F and 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=(23 x 3) x 3

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

HCF=23x3

=8x3=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

New General Mathematics, UBE Edition, chapter 1, pages 20-21

Essential Mathematics by A J S Oluwasanmi, Chapter 1, Pages 1-4

## 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 = 3x3

25= 5x5

225 = 15x15

= 3x5x3x5

= 32 x 52

9216 =96 2

=32 x 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

1. 540
2. 252

Solution

2        540

2        270

3        135

3          45

3            15        54=22 x 33x 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 3x5 =15

1.   2 252

2        126

3        63

3          21

7            7

1

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

New General Mathematics, UBE Edition, chapter 1, pages 20-21

Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 1-4

WEEKEND ASSIGNMENT

1. The lowest common multiple of 4, 6 and 8 is (a) 24 (b) 48 (c) 12 (d) 40
2. 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
3. The lowest common multiple of 4, 6 and 8 is (a) 24 (b) 48 (c) 12 (d) 40
4. The H.C.F. of 8, 24 and 36 is ___ (a) 6 (b) 4 (c) 18 (d) 20
5. The L.C.M. of 12, 16 and 24 is ___ (a) 96 (b) 48 (c) 108 (d) 24

THEORY

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

30 & 42

64 & 210