Doubt by Vanshika
Solution :
Solution :
HCF of 81 and 237 by using Euclid's Division Lemma
237>81
237>81
237 = 81×2+75 — (a)
81=75×1+6 — (b)
75=6×12+3 — (c)
6=3×2+0
∴ HCF = 3
75=6×12+3 — (c)
6=3×2+0
∴ HCF = 3
Now, using equation (c)
3 = 75-(6×12)
3 = 75-(81-75×1)×12 [Using equation (b)]
3 = 75- (81-75)×12
3 = 75-81×12+75×12
3 = -81×12 +75(1+12)
3 = -81×12+75×13
3 = 75-(6×12)
3 = 75-(81-75×1)×12 [Using equation (b)]
3 = 75- (81-75)×12
3 = 75-81×12+75×12
3 = -81×12 +75(1+12)
3 = -81×12+75×13
3 = -81×12+(237-81×2)×13 [Using equation (a)]
3 = -81×12+237×13-81×2×13
3 = -81×12+237×13-81×26
3 = -81×12+237×13-81×2×13
3 = -81×12+237×13-81×26
3 = -81(12+26)+237×13
3 = -81×38+237×13
3 = 81(-38)+237(13)
3 = -81×38+237×13
3 = 81(-38)+237(13)
This is the required linear relation between HCF, 81 and 237.
Note : The values in brackets are not unique, there could be many more possible relations.