Question

40. HCF of 272 and 148 is​

Answer 1

$\large \bold{ \underline{ \underline{ \sf \: Answer : \: \: \: }}}$

$\to$ HCF of 272 and 148 is 4

$\large \bold{ \underline{ \underline{ \sf \: Explaination : \: \: \: }}}$

$\huge \star$ By using prime factoraisation method :

$\to$ 272 = 2 × 2 × 2 × 2 × 17

$\to$ 148 = 2 × 2 × 37

Therefore ,

HCF ( 272 , 148 ) = 2 × 2 = 4

$\huge \star$ By using Euclid's algorithm :

$\to$ Since , 272 > 148 , we apply the division lemma to 272 and 148 , to get

272 = 148 × 1 + 124

$\to$ Since the remainder 124 ≠ 0 , we apply the division lemma to 148 and 124 , to get

148 = 124 × 1 + 24

$\to$ We consider the new divisor 124 and the new remainder 24 , and apply the division lemma to get

124 = 24 × 5 + 4

$\to$ We consider the new divisor 24 and the new remainder 4 , and apply the division lemma to get

24 = 4 × 6 + 0

$\to$ The remainder has now become zero , so our procedure stops , since the divisor at this stage is 4 , the HCF of 272 and 148 is 4