Question

If the $HCF$ of $210$ and $55$ is expressible in the form $210\times 5+55y,$ find $y.$

Answer 1

Let us first find the $HCF$of $210$ and $55$.

Applying Euclids division lemma on $210$ and $55$, we get
$210=55\times 3+45$

Since the remainder $45\ne 0$. So, we now apply division lemma on the divisor $55$ and the remainder $45$ to get
$55=45\times 1+10$

We consider the divisor $45$ and the remainder $10$ and apply division lemma to get
$45=4\times 10+5$

We consider the divisor $10$ and the remainder $5$ and apply division lemma to get
$10=5\times 2+0$

We observe that the remainder at this stage is zero. So, that last divisor i.e.

$5$ is the $HCF$ of $210$ and $55$.

$\therefore 5=210\times 5+55y$

$\Rightarrow 55y=5-210\times 5=5-1050$
$\Rightarrow 55y=-1045$

$\Rightarrow y=\frac{-1045}{55}=-19$