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Question

# Find the HCF of 81 and 237 and express it as a linear combination of 81 and 237.

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Solution

## Given integers are 81 and 237 such that 81<237.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 237.To represent the HCF as a linear combination of the given two numbers, we start from the last but one step and successively eliminate the previous remainder as follows: From (iii), we have3=75−6×12⇒3=75−(81−75×1)×12⇒3=75−12×81+12×75⇒3=13×75−12×81⇒3=13×(237−81×2)−12×81⇒3=13×237−26×81−12×81⇒3=13×237−38×81⇒3=237x+81y, where x=13 and y=−38.Now the HCF (say d) of two positive integers a and b can be expressed as a linear combination of a and b i.e., d=xa+yb for some integers x and y. Also, this representation is not unique. Because,d=xa+yb⇒d=xa+yb+ab−ab⇒d=(x+b)a+(y−a)bIn the above example, we had3=13×237−38×81⇒3=13×237−38×81+237×81−237×81⇒3=(13×237+237×81)+(−38×81−237×81)⇒3=(13+81)×237+(−38−237)×81⇒3=94×237−275×81⇒3=94×237+(−275)×81

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