multiples of a number - valley view english school...2. find the h.c.f of the following pairs of...
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GRADE : VI
SUBJECT: Maths
READING SECTION: Chapter:2 Properties of whole numbers
Multiples of a number
7 2
7 × 𝟏 = 7 2 × 1 = 2
7 × 2 = 14 2 × 2 = 4
7 × 3 = 21 2 × 3 = 6
7, 14, 21, 28…….. 2, 4, 6 ………. are multiple of 2
are the multiples of 7
Even
numbers
Points to Remember:
1) 1 is a factor of every number
1 x 6 = 6
1 x any number = number itself
any number ÷ 1 = number itself
8 ÷ 1 = 8
2) Every number is a factor of itself.
6 x 1 = 6
any number x 1 = number itself
any number ÷ itself = 1
8 ÷ 8 = 1
3) The factor of a number is always less than or
equal to the number.
Factor of 12: 1, 2, 3, 4, 6 and 12
1 < 12
2 < 12
3 < 12
4 < 12
6 < 12
12= 12
Factors of 10, 12 and 32 Multiples of 10, 12 and 32
Here,
10 = 1, 2, 5, 10 are factors of 10.
12 = 1, 2, 3, 4, 6, 12 are factors of 12
32 = 1, 2, 4, 8, 16, 32are factors of 32
Here,
Multiples of 10 ; 10, 20, 30, 40, 50…….are
multiples of 10.
Multiples of 12; 12, 24, 36, 48, 60 ……. are
the multiple of the 12
Multiples of 32: 32, 64, 96, 128, 160……. are
the multiple of 32
Factors and multiples:
Prime factorization for the given numbers.
36 45 70
36 = 2×2×3×3 45=3×3×5 70= 2×5×7
Note: To get prime factors ,we divide by only prime numbers like 2,3,5 ,7 and so on
1. Write all the possible factors of these numbers.
a) 16 b) 35 c) 64 d) 18 e) 140
2. Write the first five multiples of these numbers.
a) 7 b) 9 c) 11 d) 15 e) 25
3. Factorize the following numbers using prime factorization process.
a) 18 b) 24 c) 72 d) 100 e) 150
Exercise:1
Worked Out Example:
HCF = Product of their common prime
factors HCF stands for Highest Common Factor
Three methods of finding HCF
Writing all the
possible factors Prime factorization
method Division method
Writing all the possible factors
a) 16, 20
Factors of 16= 1, 2, 4, 8, 16
Factors of 20= 1, 2, 4, 5, 10, 20
Here, common factors= 1, 2, 4
(among these common factors, 4 is
the highest )
Therefore H.C.F of 16,20=4
b) 14, 28,42
Factors of 14= 1, 2, 4, 7, 14
Factors of 28= 1, 2, 4, 7, 14, 28
Factors of 42= 1, 2, 3, 6, 7, 14, 21,
42
Here, common factors = 1,2,7,14
(among these common factors, 14 is
the highest )
Therefore H.C.F of 14,28 42= 14
Method One
Prime factorisation method
a)16,20 b) 14, 28, 42
H.C.F= 2 × 2 H.C.F= 2× 7
= 4 = 14
Method Two
Method Three
Divide the bigger number by the smaller one, divide each
divisor by corresponding remainder till the remainder comes to
zero
a) 16, 20
Note: The last divisor is the H.C.F.
HCF = 4
Step 1: Divide the bigger number by smaller one, divide each
divisor by corresponding remainder till remainder comes zero.
Here HCF
of 60 and
96 = 12
Step 2: Again, divide the third number by the obtained HCF and
follow the previous step till the remainder comes zero.
Hence HCF of 60, 96 and 150 = 6.
1.Write all the possible factors of the following pairs of numbers , circle the common factors and
select the H.C.F
a) 8 , 12 b) 16, 24 c) 20, 30 d) 24 , 36 e) 15, 30, 45
2. Find the H.C.F of the following pairs of numbers by prime factorization method.
a) 12, 16 b) 15, 20 c) 24, 30 d) 44, 66
e) 12, 18, 24 f) 27, 45, 63 g) 90, 150, 180
3. Find the H.C.F of the following pairs of numbers by division method.
a) 16, 24 b) 15, 20 c) 26, 39 d) 54, 90
e) 80, 120 f) 27, 45, 63 g) 80, 120, 200
Exercise: 2
LCM stands for Lowest Common Multiple
LCM = Product of their common prime factors and
remaining factors
Three methods of finding LCM
Writing the first
ten multiples Division
process. Prime factorization
method
Let’s find L.C.M of : a) 8, 12 b) 6, 10, 15
Writing the first ten multiples
a) 8, 12
Multiples of 8= 8, 16, 24, 32, 40, 48,
56, 64, 72, 80
Multiples of 12= 12, 24, 36, 48, 60, 72,
84, 96 ,108,
Here, common multiples are 24, 48 and
72
(among these common multiples 24 is
the lowest)
Therefore L.C.M of 8, 12= 24
b) 6, 10, 15
Multiples of 6= 6, 12, 18, 24, 30,
36, 42, 48, 54, 60
Multiples of 10= 10, 20, 30, 40, 50,
60, 70, 80, 90, 100 , 120
Multiples of 15= 15, 30, 45, 60, 75,
90, 105, 120, 135, 150
Here common multiples are 30, 60
(among these common multiples 30
is the lowest)
Therefore L.C.M of 6, 10, 15 = 30
Method One
Method Two
Division Process
a) 8, 12
L.C.M = 2 × 2 × 2 × 3
= 24
b) 6, 10, 15
L.C.M = 2 × 3 × 5
= 30
Prime factorization method
a) 8, 12
Method Three
8 = 2 x 2 x 2
12= 2 x 2 x 3
L.C.M = 2 x 2 x 2 x 3
= 24
Take 2 x 2 from first
and second number
then, multiply the
remaining number first
and remaining number
of second
b) 24, 36, 60
24 = 2 x 2 x 2 x 3
36 = 2 x 2 x 3 x 3
60 = 2 x 2 x 3 x 5
L.C.M = 2 x 2 x 2 x 3 x 3 x 5
= 360
here, among 2x2x2 and 2x2, select
the highest value i.e 2x2x2.
Similarly among 3 and 3x3 select
3x3
5 is only lef so take it, because there
is no option for 5.
1.Write the first ten multiples of the following pairs of numbers and select the L.C.M
a) 2, 4 b) 6, 8 c) 7, 14 d) 10 , 20 e) 4, 6,
2. Find the L.C.M of the following pairs of numbers by division method.
a) 8, 10 b) 9, 15 c) 16, 24 d) 22, 33
e) 8, 12, 16 f) 18, 24, 36 g) 40, 45, 60
3. Find the L.C.M of the following pairs of numbers by prime factorisation method.
a) 12, 15 b) 30, 45 c) 32, 48 d) 16, 24, 32
e) 25, 50, 75 f) 36, 48, 60 g) 120, 144, 160
Exercise: 3
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