Question

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

Answer 1

45=9×5 and 27=9×3

⟹hcf(45, 27)=9=d.

d=27x+45y

⟹9=27x+45y

⟹27x=9−45y

⟹27x=−27y+9−18y

⟹ 27x27=−27y 27+9−18y27

⟹x=−y+9(1−2y)9×3

⟹x=−y+1−2y3.

t=1−2y3, t∈Z,  

⟹3t=1−2y

⟹2y=1−3t

⟹2y=−2t+1−t

⟹ 2y2=−2t  2+1−t2

⟹y=−t+1−t2.

u=1−t2

⟹2u=1−t, u∈Z,  

⟹t=1−2u

⟹y=−(1−2u)+u

=−1+2u+u  

=−1+3u.

x=−y+1−2y3

⟹x=−(−1+3u)+1−2(−1+3u)3

=1−3u+1+2−6u3

=1−3u+3−6u3

=1−3u+3(1−2u)3

=1−3u+1−2u

=2−5u.

∴ (x, y)=(2−5u, −1+3u), u∈Z.