Question

In a two digit number the sum of the two digits is $8$. If $18$ is added to the number its digits are reversed. Find the number.

Answer 1

Let the digit at unit's place be '$x$'
Then the digit at ten's place $=8-x$ (sum of the two digits is $8$)
Therefore number will be $10(8-x)+x=80-10x+x=80-9x.......\left(i\right)$
Now ,number obtained by reversing the digits $=10\times \left(x\right)+(8-x)$
$=10x+8-x$

$=9x+8$
It is given that if $18$ is added to the number its digits are reversed
$\therefore$ The number $+18=$ Number obtained by reversing the digits
$\Rightarrow$ $(80-9x)+18=9x+8$
$98-9x=9x+8$
$98-8=9x+9x$
$90=18x$
$x=\frac{90}{18}=5$
By substituting the value of $x$ in equation $\left(i\right)$ we get the number
$\therefore$ Number $=80-9\times 5=80-45=35$