Question

The sum of $2$ digit number and the number obtained by reversing the order of the digit is $165$. If the digit differs by $3$, find the number, when ten's digit is bigger than the unit's digit.

Answer 1

Step 1:Determine the original number

Given information,

The number is two-digit $=10\mathrm{x}+\mathrm{y}$ (assume)

The sums of the number and its reverse $=165$

Difference between digits $=3$

According to the problem,

$(10\mathrm{x}+\mathrm{y})+(10\mathrm{y}+\mathrm{x})=165$
$\Rightarrow$ $11\mathrm{y}+11\mathrm{x}=165$
$\Rightarrow$ $11(\mathrm{x}+\mathrm{y})=165$
$\Rightarrow$ $\mathrm{x}+\mathrm{y}=\frac{165}{11}$

$\Rightarrow$ $\mathrm{x}+\mathrm{y}=15\dots ..\left(1\right)$
$\Rightarrow$ $\mathrm{x}-\mathrm{y}=3\dots ...\left(2\right)$

Step 2:Solve the equation:

On solving equations $\left(1\right)$ and $\left(2\right)$

$\mathrm{x}+\mathrm{y}+\mathrm{x}-\mathrm{y}=15+3$
$$\begin{gathered} \Rightarrow 2\mathrm{x}=18 \\ \Rightarrow \mathrm{x}=\frac{18}{2} \\ \Rightarrow \mathrm{x}=9 \end{gathered}$$
$\Rightarrow$ $\mathrm{x}+\mathrm{y}=15$
$\Rightarrow$ $9+y=15$
$\Rightarrow$ $$\begin{gathered} \mathrm{y}=15-9 \\ \mathrm{y}=6 \end{gathered}$$
$\Rightarrow \mathrm{x}=9,\mathrm{y}=6$
Therefore, the original number is $96$.