Question

find he hcf of 963and 657 and express it as linear equtions

Answer 1

Using Euclid’s Division Lemma 
a = bq+r , o ≤ r < b 
963 = 657×1 + 306 -------(1)
657 = 306×2 + 45 ---------(2)
306 = 45×6+36 ------------(3)
45 = 36×1+9 --------------(4)
36 = 9×4+0 
∴ HCF (657, 963) = 9

From eq. (4)...........
45 = 36×1+9
9 = 45-36×1
9 = 45-(306-45×7)×1 [from eq.(3)]
9 = 45- 306+45×7 ×1
= 45×7 -306
= (657-306×2)×7-306[from eq(2)]
= 657×7-306×14-306
= 657×7-306×15
= 657×7-(963-657×1)×15[from eq(1)]
= 657×7-963×15+657×15
= 657×22-963×15
9 = 657x - 963y (where x=22, y=15)

Hope It helped You.............