Question

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

Answer 1

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.