Question

The sum of a two-digit number and the number formed by interchanging the digits is $132$. If $12$ is added to the number, the number becomes $5$ times the sum of the digits. Find the number.

Answer 1

Step 1: Suppose the number

Let the tens digit of the number be $x$, and the unit digit be $y$.

Therefore, the original number $=10x+y$.

The number is obtained by interchanging the digits$=x+10y$.

According to the question,

The sum of a two-digit number and the number formed by interchanging the digits is $132$.

So the equation becomes:

$\begin{array}{rcl}10x+y+10y+x& =& 132\\ 11x+11y& =& 132\\ x+y& =& 12.....\left(1\right)\end{array}$

If $12$ is added to the number, the number becomes $5$ times the sum of the digits.

So the equation becomes:

$\begin{array}{rcl}10x+y+12& =& 5(x+y)\\ 10x+y+12& =& 5x+5y\\ 5x-4y& =& -12...........\left(2\right)\end{array}$

Step 2: Solve equations (1) and (2) by the method of substitution

Substitute $x=12-y$ in equation (2) to get:

$\begin{array}{rcl}5(12-y)-4y& =& -12\\ 60-5y-4y& =& -12\\ -9y& =& -72\\ y& =& \frac{72}{9}\\ \mathit{y}& \mathbf{=}& \mathbf{8}\end{array}$

Substitute $y=8\text{in}x=12-y$ to get:

$\begin{array}{rcl}x& =& 12-8\\ \mathit{x}& \mathbf{=}& \mathbf{4}\end{array}$

Now substitute the value of $x\text{and}y$ to get the desired number:

$\begin{array}{rcl}\text{Number}& =& 10\left(4\right)+\left(8\right)\\ & =& 48\end{array}$

Hence, the required number is $48$.