Question

If $d$ is $HCF$ of $45$ & $27$. Find $x$ satisfying $d=27x-45$

Answer 1

Given $d$ is $HCF$ of $45$ & $27$

$\begin{array}{c}27\left)45\right(1\\ \underset{–}{27}\\ 18\left)27\right(1\\ \underset{–}{18}\\ 9\left)18\right(2\\ \underset{–}{18}\\ \times \times \end{array}$

$HCF$ is $d=9$

$d=27x-45$

$9=27x-45$

$9+45=27x$

$54=2x$

$x=2$