Question

In the HCF of $657$ and $963$ is expressible in the form $657x+963$ $\times$ $-15$ ,find $x$.

Answer 1

$\begin{array}{c}963=657\times 1+306\\ 657=306\times 2+45\\ 306=45\times 6+36\\ 45=36\times 1+9\\ 36=9\times 4\end{array}$

HCF $=9$

$\begin{array}{c}657x+963\times -15=9\\ 657x-14445=9\\ x=\frac{14454}{657}\\ x=22\end{array}$