Find HCF of 81 and 237 and express it as a linearcombination of 81 and

1.	Find HCF of 81 and 237 and express it as a linearcombination of 81 and 237 ie., HCF (81, 237)81x + 237y for some x and y.
Answer» By Euclid's Division Algorithm, 237=81(2)+(75) 81=75(1) + (6) 75=6(12)+(3) 6=3(2)+(0) HCF =3 Expressing it in the form of 237x+81y=HCF 3=75-6(12) { From 2nd last step} 3=75-(81-75)(12) {Substituting} 3=75-(81×12-75×12) 3=75-81×12+75×12 3=75(13)-81(12) 3=(237-81×2)(13)-81(12) 3=237(13)-81(38) 3 = 237 (13) + 81 (-38) {we need an expression in the form 237x + 81y }Therefore, x =13 , y =- 38

Find HCF of 81 and 237 and express it as a linearcombination of 81 and 237 ie., HCF (81, 237)81x + 237y for some x and y.

Answer»

By Euclid's Division Algorithm,

237=81(2)+(75)

81=75(1) + (6)

75=6(12)+(3)

6=3(2)+(0)

HCF =3

Expressing it in the form of 237x+81y=HCF

3=75-6(12) { From 2nd last step}

3=75-(81-75)(12) {Substituting}

3=75-(81×12-75×12)

3=75-81×12+75×12

3=75(13)-81(12)

3=(237-81×2)(13)-81(12)

3=237(13)-81(38)

3 = 237 (13) + 81 (-38) {we need an expression in the form 237x + 81y }Therefore, x =13 , y =- 38