Back to: MATHEMATICS JSS 2
Welcome to Class !!
We are eager to have you join us !!
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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In our next class, we will be talking about Fractions as Ratios, Decimals and Percentages. We are very much eager to meet you there.
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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