Question

If the sum of two numbers is 1215 and their HCF is 81 then how many pairs of such numbers are possible?
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Answer 1

Given:  The sum of two numbers is$1215$ and their HCF is$81$.

We have to find the number of pairs of such numbers is possible.

We are solving in the following way:

We have,

The sum of two numbers is$1215$ and their HCF is$81$

$\text { Let the two numbers be } x \text { and } y \text {. }$

Now,

$\begin{array}{l}=>81 \mathrm{x}+81 \mathrm{y}=1215 \quad \ldots[\text { sum of two numbers are 1215 }] \\\Rightarrow 81(\mathrm{x}+\mathrm{y})=1215 \\\Rightarrow \mathrm{x}+\mathrm{y}=15\end{array}$

So,

$\text { For, } x=1, y=14, \text { the numbers are } 1 \times 81+14 \times 81=81+1134=1215\\\\\text { For, } x=7, y=8 \text {, the numbers are } 7 \times 81+8 \times 81=567+648=1215\\\\\text { For, } x=2, y=13, \text { the numbers are } 2 \times 81+13 \times 81=162+1053=1215\\\\\text { For, } x=4, y=11, \text { the numbers are } 4 \times 81+11 \times 81=324+891=1215$

Hence, the numbers of such pairs are $4$.