Question

2. Find g.c.d. of the following by Euclid's
algorithm
(1) 12576 and 4052 (2) 81 and 237​

Answer 1

Applying division lemma to 81 and 237, we get

Applying division lemma to 81 and 237, we get237=81×2+75 

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  We consider the new divisor 75 and the new remainder 6 and apply division lemma to get

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  We consider the new divisor 75 and the new remainder 6 and apply division lemma to get75=6×12+3  

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  We consider the new divisor 75 and the new remainder 6 and apply division lemma to get75=6×12+3  We consider the new divisor 6 and the new remainder 3 and apply division lemma to get

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  We consider the new divisor 75 and the new remainder 6 and apply division lemma to get75=6×12+3  We consider the new divisor 6 and the new remainder 3 and apply division lemma to get6=3×2+0  

Applying division lemma to 81 and 237, we get237=81×2+75 Since the remainder 75=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get81=75×1+6  We consider the new divisor 75 and the new remainder 6 and apply division lemma to get75=6×12+3  We consider the new divisor 6 and the new remainder 3 and apply division lemma to get6=3×2+0  The remainder at this stage is zero. So, the divisor at this stage or the remainder at the earlier stage i.e. 3 is the HCF of 81 and 

Answer 2

Answer:

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