Question

The sum of a two-digit number and the number obtained by reversing its digits is $99$ and if digits differ by 3.Find the number.

Answer 1

We have,

Case (I)-

$\Rightarrow$$1$ unit place $=x$

$\Rightarrow$$10$ unit place $=y$

$\therefore 10y+x$

Case (II)

$10$ units place $=x$

$1$ unit place $=y$

$\therefore y+10x$

$10y+x+y+10x=99$

$11y+11x=99$

$11(y+x)=99---\left(I\right)$

$x+y=9$

$\therefore x-y=3----\left(II\right)$

put the value $\left(i\right)-\left(ii\right)$

$2x=12$

$\Rightarrow$$x=6$

$\therefore$ sum of two-digit number

$36$ or $63$

Hence this is the answer.