Math, asked by manikantamshetty12, 6 months ago

find HCF of 81 and 237 using Euclid division​

Answers

Answered by Anonymous
1

Answer:

Step-by-step explanation:

Since the remainder 75

=0. So, consider the divisor 81 and the remainder 75 arndapply division lemma to get

81=75×1+6

We consider the new divisor 75 and the new remainder 6 and apply division lemma to get

75=6×12+3

We consider the new divisor 6 and the new remainder 3 and apply division lemma to get

6=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 have

3=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)b

In the above example, we had

3=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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