Question

Hcf of 18,54,81 by repeated division method

Answer 1

Answer:

9 is the HCF

mark me as brainlist

Answer 2

First find all factors of the given numbers individually.

Factors of 18 = 1, 2, 3, 6, 9, 18

Factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24

Factors of 42 = 1, 2, 3, 6, 7, 14, 21, 42

Therefore, common factors of 18, 24 and 42 are 1, 2, 3 and 6.

HCF of 18, 24 and 42 is 6.

Prime factorization method

Let’s find the HCF of 27 and 45.

First we have to find the factors of 27.

Write a pair of factors.

27 = 3 x 9

Now further factorize the composite factor 9 as 3 and 3

Therefore, 27 as a product of its prime factors is written as

27 = 3 x 3 x 3

Let’s find the factors of 45.

Write a pair of factors.

45 = 5 x 9

Further factorize the composite factor 9 as 3 and 3.

Therefore, 45 as a product of its prime factors is written as

45 = 5 x 3 x 3

Multiply all the factors which appear in both the list.

27 = 3 x 3 x 3

45 = 3 x 3 x 5

i.e 3 x 3 = 9

Therefore, HCF of 27 and 45 is 9.

Question: Find the HCF of the following numbers. 18, 54, 81

Solution: Let’s find the HCF by prime factorization method.

18 = 2 x 3 x 3

54 = 2 x 3 x 3 x 3

81 = 3 x 3 x 3 x 3

Multiply all the factors which appear in both the list. i.e 3 x 3 = 9

The HCF of 18, 54 and 81 is 9.