Question

D is the hcf of 45 and 27 find x and y satisfying d=27x+45y

Answer 1

Hcf of 45 and 27

45 = (27 x 1) + 18
27 = (18 x 1) + 9

18 = (9 x 2) + 0

Therefore, HCF = 9

9 = 27 – (18 x 1)
= 27 – [45 – (27 x 1)] x 1
[since 18 = 45 – (27 x 1)]

= 27 – [45 x 1 – 27 x 1 x 1]

= 27 – (45 x 1) + (27 x 1)

= 27 + (27 x 1) – (45 x 1)

= 27{1 + 1} – (45 x 1)

= (27 x 2) – (45 x 1)

HCF of 45 and 27 in the form of 27x + 45 is (27 x 2) + (45 x -1).

Hence, the values of x and y are 2 and -1.

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