Question

find hcf of 81 and 237 and express it as a linear condition of 81 and 237

Answer 1

237281,we apply Euclid division lemma
237=81×2+75.....(1)
81=75×1+6......(2)
75=6×12+3...(3)
6=3×2+0
here, divisor is 3
HCF of 237 and 81 is 3
3=75-6×12... (from 3)
75-(81-75×1)×12( from 2)
75-81×12+75×12
75×13-81 ×12
(237-81×2)×13-81×12( from 1)
237×13 -81×26-81×12
237×13-81×38
237×13+81(-38)
81(-38)+237(13)
81x+237y=81(-38) + 237(13)
now, x=-38 and y=13