Find the hcf of 272 and 148 and express it as 272p+148q

1.	Find the hcf of 272 and 148 and express it as 272p+148q
Answer» How do we get the value of p and q \xa0272 = 2 × 2 × 2 × 2 × 17\xa0148 = 2 × 2 × 37Therefore ,HCF ( 272 , 148 ) = 2 × 2 = 4\xa0By using Euclid\'s algorithm :\xa0Since , 272 > 148 , we apply the division lemma to 272 and 148 , to get272 = 148 × 1 + 124\xa0Since the remainder 124 ≠ 0 , we apply the division lemma to 148 and 124 , to get148 = 124 × 1 + 24\xa0We consider the new divisor 124 and the new remainder 24 , and apply the division lemma to get124 = 24 × 5 + 4\xa0We consider the new divisor 24 and the new remainder 4 , and apply the division lemma to get24 = 4 × 6 + 0\xa0The 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 P=(6) , q=(-11)

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