Question

Find HCF of 81 and 237 express HCF in the form of 237*p+81*q

Answer 1

H.C.F. of 81 and 237 , by using Euclid's division lemma , As

237 = 81 × 2 + 75 ----------- ( 1 )

81 = 75 × 1 + 6 --------- ( 2 )

75 = 6 × 12 + 3 ---------- ( 3 )

6 = 3 × 2 + 0

Hence

H.C.F. of 81 and 237 is = 3 ( Ans )

From equation 3 , we get

3 = 75 - 6 × 12

And from 2 , we get

6 = 81 - 75 × 1

SO,

3 = 75 - ( 81 - 75 × 1 ) × 12

3 = 75 - ( 81 ×12 - 75 × 12 )

3 = 75 × 13 - 81 ×12

From equation 1 we get

75 = 237 - 81× 2

So,

3 = ( 237 - 81 × 2 ) × 13 - 81 × 12

3 = ( 237 × 13 - 81 × 26 ) - 81 × 12

3 = 237 × 13 - 81 × 38

3 = 237 × 13 + 81 × - 38 ( Ans )

Answer 2

Answer:

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