Question

The sum of the digits of a two digit number is $12$.

The number obtained by interchanging its digits exceed the given number by $18$.

Find the number.

Answer 1

Step 1: Let the tens and the units digits of the required number be $x$ and $y$ respectively.

The two-digit number $=(10x+y)$

Given, the sum of the digits of the number is equal to $12$.

$x+y=12..........\left(i\right)$

Number obtained on reversing its digits $=(10y+x)$

According to the question:

$$\begin{gathered} \begin{array}{r}(10y+x)-(10x+y)=18\\ 10y+x-10x-y=18\\ 9y-9x=18\end{array} \\ y-x=2..........\left(ii\right) \end{gathered}$$

Step 2: Find the required numbers

On adding $\left(i\right)$ and $\left(ii\right)$

We get.

$\begin{array}{rcl}2y& =& 14\\ y& =& 7\end{array}$

Substitue $y=7$ in $\left(i\right)$, we get

$\begin{array}{rcl}x+7& =& 12\\ x& =& 12-7\\ x& =& 5\end{array}$

The two-digit number $=(10x+y)$

The two-digit number $=(10\times 5)+7$

The two-digit number $=50+7$

The two-digit number $=57$

Hence, the required number is $57$.