Find HCF of 81 and 273 and express it as a linear combinationof 81 and 273i.e.,HCF(81 and 273)=81x+273y for some x and y
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By Euclid' 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=(23-81*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
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=(23-81*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
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