Question

The sum of a two digit number and the number obtained by reversing the digits is $66$. If the digits of the number differ by $2$, find the number. How many such numbers are there?

Answer 1

Let unit be $x$
The tens place digit will be $x+2$
so, the number =$\left(x\right)\times 1+(x+2)\times 10$
=$x+10x+20$
=$11x+20$
If digits are reversed
then the number obtained=$x+2+10x$
But given
$\begin{array}{c}11x+20+11x+2=66\\ 22x=66-22\\ 22x=44\\ x=2\end{array}$
therefore required number
$\begin{array}{c}=11x+20\\ =11\times 2+20\\ =22+20\\ =42\end{array}$
When we reverse the number 42, we get 24 and again this number also satisfies the given condition.
Hence, there are two such numbers.