Find the HCF of the following numbers by prime factorisation and continued division method
1)18,27,36
2)106,159,265
3)10,35,40
4)32,64,96,128
I want this
Answers
Answer:
(i) 18,27,36
The three numbers can be represented as the product of their prime factors, as 3
2
×2, 3
3
and 3
2
×2
2
respectively.
The product of the lowest powers of the common factors among the three numbers is 2
1
×3
2
∴ HCF=2×3
2
=18
(ii)106,159,165
The three numbers can be represented as the product of their prime factors, as 2×53,3×53 and 5×53 respectively.
The product of the lowest powers of the common factors among the three numbers is 53
1
∴ HCF=53
(iii)10,35,40
The three numbers can be represented as the product of their prime factors, as 5×2,5×7, and 2
3
×5 respectively.
The product of the lowest powers of the common factors among the three numbers is 5
1
∴ HCF=5
(iv)32,64,96,128
The three numbers can be represented as the product of their prime factors, as 2
5
,2
6
,2
5
×3 and 2
7
respectively.
The product of the lowest powers of the common factors among the three numbers is 2
5
∴ HCF=2
5
=32
Step-by-step explanation:
answer is in picture
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answer of the other question is also herewith
