Question

The sum of a two digit number and the number obtained by reversing the order of its digits is $99$. If the digits differ by $3$, find the number.

Answer 1

Let the $2$ digits be $x,y.$ respectively

$2$ digit number $=10x+y$

$2$ digit number obtained by ususing the digit $=10y+x$

$\therefore 10x+y+10y+x=99\Rightarrow 11x+11y=99\Rightarrow x+y=9x-y=3$

On adding : $2x=12$

$x=6$

putting $x$ in $⟶\left(i\right)$

$6+y=9$

$y=3$

$\therefore$ Number $=10x+y=63$

or , $10y+x=36.$