The sum of the digit of a 2 digit number is 6 . On reversing it's digits, the number is 18 less than the original number find the number

Given :

Sum of the digits of a two-digit number$=6$.

On reversing it's digits, the number is 18 less than the original number.

To do :

We have to find the two-digit number.

Solution :

Let the unit digit be y and tens digit be x.

The number formed $= 10x + y$

The reverse number $= 10y + x$

Therefore,

$x + y = 6$.........................(i)

$10y + x = (10x + y) - 18$

$ 10y - y + x -10x = -18$

$9y - 9x =-18$

$y-x =-2$

$y=x-2$.................(ii)

Substitute (ii) in (i)

$x + (x-2) = 6$

$2x -2 = 6$

$2x = 8$

$x = \frac{8}{2} = 4$

$y= 4-2 = 2$

So, $x = 4, y= 2$

Therefore, the original number is 42.