Question

# If HCF of $210$ and $55$ is of the form $(210)(5) + 55y$, then the value of $y$ is:$(a){\text{ }} - 19$$(b){\text{ }} - 18$$(c){\text{ }}5$$(d){\text{ }}55$

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Hint: Factorise the given numbers carefully. Also keep in mind that the value of $y$ can be negative as well.

We have the given numbers as:
$210$ and $55$
Now, these can be factorized as
$210 = 2 \times 3 \times 5 \times 7$
$55 = 5 \times 11$
This shows that the HCF of $210$ and $55$ comes out to be $5$.
Since, it is given in the question that the HCF of $210$ and $55$ is of the form $(210)(5) + 55y$
That is,
$5 = (210)(5) + 55y$
On further evaluation, we get
$\Rightarrow 1050 + 55y = 5$
$\Rightarrow 55y = - 1050 + 5$
$\Rightarrow 55y = - 1045$
$\Rightarrow y = \dfrac{{ - 1045}}{{55}}$
$\therefore y = - 19$
So, the required solution is $(a){\text{ }} - 19$.

Note: To solve these types of questions, find out the HCF of the given numbers and substitute it in the given equation to obtain the required solution.