Question

If d is the HCF of 45 and 27 ,find d=27x + 45y.​

Answer 1

$\huge{\mathfrak{Answer}}$

x = 2 and y = -1

___________________________________

$\huge{\mathfrak{Solution}}$

HCF of 45 and 27 can be calculated as:

45 = 27 × 1 + 18

27 = 18 × 1 + 9

18 = 9 × 2 + 0

Therefore, HCF = 9

ATQ,

$\sf d = 27x + 45y \\ \\ \\ \sf{ From \: above \: steps \: we \: have } \\ \sf9 = 27 - 18 \times 1 \\ \sf \: \: \: = 27 - (45 - 27 \times 1) \times 1 \\ \: \: \: \sf = 27 \times 2 + 45 \times ( - 1) \\ \\ \sf{Comparing \: this \: with \: given \: situation \: we \: get} \\ \sf x = 2 \: and \: y = - 1$

____________________________________

Answer 2

HCF of 45 and 27

=>45 = 27 x 1 + 18

=> 27 = 18 x 1 + 9

=> 18 = 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)

Hope this will help you ☺