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If did HCF of 45,27. Find x and y satisfying d=27x + 45y |
| Answer» HCF of 45 and 27 -45 = 27 x 1 + 1827 = 18 x 1 + 918 = 9 x 2 + 0 So, HCF (45, 27) = 9 d = 9 = 27x + 45y Make a linear combination- 9=27-18×1 =27-(45-27×1)×1 (18=45-27×1) =27-45×1+27×1 =2×27-1×45 =27x+45y (x=2,y=-1) | |