Question

find the HCF of 45 and 27 the HCF is expressed as 27 X + 45 Y find X and Y​

Answer 1

Applying Euclid's division lemma to 27 and 45, we get

45=27×1+18 ...(1)

27=18×1+9 ...(2)

18=9×2+0 ...(3)

Since the remainder is zero, therefore, last divisor 9 is the HCF of 27 and 45.

From (2), we get

9=27−18×1

=27−(45−27×1)×1 [using(1)]

=27−45×1+27×1×1

=27−45+27

=54−45

⇒9=27×2−45×1 ...(4)

Comparing (4) with d=27x+45y, we get

d=9,x=2 and y=−1