Question

1.
(a) 64, 32
(d) 13, 52, 260,-
(c) 56, 84
(b) 45, 135
(e) 140, 160, 240
Find the H.C.F. of the following pairs of numbers, by using prime factorisation method :​

Answer 1

Answers

HCF of

(a) 64, 32

To Find the HCF of 64 and 32 , let us do the prime factorization of the same.

$\implies$

64 = 2 × 2 × 2 × 2 × 2 × 2

32 = 2 × 2 × 2 × 2 × 2

$\implies$HCF of 64 and 32 = 2 × 2 × 2 × 2 × 2 = 32.

(b) 45, 135

To Find the HCF of 45 and 135 , let us do the prime factorization of the same.

$\implies$

45 = 3 × 3 × 5

135 = 3 × 3 × 3 × 5

$\implies$  HCF of 45 and 135 = 3 × 3 × 5 = 45.

(c) 56, 84

To Find the HCF of 56 and 84 , let us do the prime factorization of the same.

$\implies$

56 = 2 × 2 × 2 × 7

84 = 2 × 2 × 3 × 7

$\implies$ HCF of 56 and 84 = 2 × 2 × 7 = 28.

d) 13, 52, 260

To Find the HCF of 13, 52, 260  , let us do the prime factorization of the same.

$\implies$

13 = 13

52 = 2 × 2 × 13

260 = 2 × 2 × 5 × 13

$\implies$ HCF of 13, 52 and 260 = 13

(e) 140, 160, 240

To Find the HCF of  140, 160 and 240 , let us do the prime factorization of the same.

$\implies$

140 = 2 × 2 × 5 × 7

160 = 2 × 2 × 2 × 2 × 2 × 5

240 = 2 × 2 × 2 × 2 × 3 × 5

$\implies$ HCF of  140, 160 and 240 =  2 × 2 × 5 = 20.