Back to: MATHEMATICS JSS 2
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In today’s Mathematics class, We will be discussing LCM, HCF and Perfect Squares. We hope you enjoy the class!
Highest Common Factors
A 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:
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
Classwork
Find the HCF of:
(1) 28 and 42
(2) 504 and 588
(3) Find the LCM of 84 & 210
READING ASSIGNMENT
New General Mathematics, UBE Edition, chapter 1, pages 2021
Essential Mathematics by A J S Oluwasanmi, Chapter 1, Pages 14
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)
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
READING ASSIGNMENT
New General Mathematics, UBE Edition, chapter 1, pages 2021
Essential Mathematics by A J S Oluwasanmi, Chapter 1, pages 14
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

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very interesting
I love your class but what about indices am finding it difficult to understand pls give a brief explanation thank you.
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For the first class work for this week
28=2*2*7=28
42=3*2*7=42
=2*7=14.
2.)504 & 588
504=2*2*2*3*3*7=504
588=2*3*7*7
=2*3*7=42.
3) Lcm of 84 & 210
84=2*3*3*7
210=2*3*5*7
2*3*7=42