The sum of two numbers is $1215$ and their HCF is $81$. How many pairs of such numbers can be formed?

The correct option is C $\left(4\right)$
It is given that the sum of two numbers is $1215$ and their H.C.F is $81$.
Let the two numbers be $x$ and $y$.
Now,
$81x+81y=1215$ ...[sum of two numbers are $1215$]
$\Rightarrow 81(x+y)=1215$
$\Rightarrow x+y=15$ .......(1)

Case I: For, $x=1$ in equation(1), we get
$y=15-1$
$\Rightarrow y=14$
Substitute the values of $x$ and $y$ in equation(1), we get
$1\times 81+14\times 81=81+1134=1215$
Thus, $(1,14)$ is one pair.
Similar way, we can verify for other pairs as follow.
Case II: For, $x=7,y=15-7=8$,
the numbers are $7\times 81+8\times 81=567+648=1215$
Case III: For, $x=2,y=15-2=13$,
the numbers are $2\times 81+13\times 81=162+1053=1215$
Case IV: For, $x=4,y=15-4=11$,
the numbers are $4\times 81+11\times 81=324+891=1215$
Therefore, the numbers of such pairs are $4$.