Question

If $HCF$ of $210$ and $55$ is expressible in the form $210\ast 5+55\ast A$, then find the value of $(2-A)$.

Answer 1

$210>55$

$210=55\times 3+45$

$55=45\times 1+10$

$45=10\times 4+5$

$10=5\times 2+0$

$\therefore HCF=5$

$5=45-\left(10\times 4\right)$

$=45-\left(55-45\times 1\right)\times 4$

$=45-\left(55\times 4\right)+\left(45\times 4\right)$

$=\left(45\times 4\right)-\left(55\times 4\right)$

$=\left\{\left(210-55\times 3\right)\times 5\right\}-\left(55\times 4\right)$

$=\left(210\times 5\right)-\left(55\times 15\right)-\left(55\times 4\right)$

$=\left(210\times 5\right)-\left(55\times 19\right)$

$=\left(210\times 5\right)+\left(55\times -19\right)$

$\therefore A=-19$

value of $\left(2-A\right)$ $=2+19=21$.