Question

If sum of two numbers is $1215$ and their $\mathrm{HCF}$ is $81$ then , How many such pairs of numbers are possible?

Answer 1

Find the possible number of pairs

Since $\mathrm{HCF}$ of both numbers is $81$ so both will be a multiple of $81$.

Let the first number be $=81\mathrm{x}$

Let the second number $=81\mathrm{y}$

From given data,

$\begin{array}{cc}& 81\mathrm{x}+81\mathrm{y}=1215\\ \Rightarrow & \mathrm{x}+\mathrm{y}=\frac{1215}{81}\\ \Rightarrow & \mathrm{x}+\mathrm{y}=15\end{array}$

Thus, when $x$ and $y$ satisfy the above condition, the given condition is met.

The possible pairs of $x$ and $y$ are,

$\left\{\left(0,15\right),\left(1,14\right),\left(2,13\right),\dots ,\left(13,2\right),\left(14,1\right),\left(15,0\right)\right\}$

Ignoring the pairs where one of them is $0$ (because if we do consider it, one of the numbers will be $0$), we have $13$ pairs of numbers.

We observe that the set is symmetric, i.e., there are elements such as $\left(a,b\right)$ and $\left(b,a\right)$.

Also ignoring those, we have $\left\{\left(1,14\right),\left(2,13\right),\dots ,\left(7,8\right)\right\}$

So the number of possible pairs are $\left\{\left(1,14\right),\left(2,13\right),\dots ,\left(7,8\right)\right\}$

Hence, the number of possible pairs are $7$.