Question

The sum of the digit in a two digits number is $9$. The number obtained by interchanging the digits exceeds the original number by $27$. Find the two digits numbers.

Answer 1

Let the two digit number be $10x+y$

Given that the sum of the digits is $9$

$x+y=9$------------------$\left(1\right))$

Given that the number obtained by interchanging the digits exceeds the given number by $27$

$10y+x=10x+y+27$

$9x-9y=-27$

taking $9$ as common

$x-y=-3$---------------------$\left(2\right)$

Adding equation $1$ and $2$

$x+y=9$

$x-y=-3$

$2x=6$

$x=3$

$3+y=9$

$y=6$

The number is $10x+y$ is $36.$