Question

The sum of the digits in a two-digit number is $6$. If we add $18$ to that number, we get a number consisting of the same digits written in the reverse order. Find the number.

Answer 1

Let the digit at ones place be $x$ and that at tens place be y.

Therefore the number is $10y+x$.

After reversing, the number formed is $10x+y$.

According to question

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

$9x-9y=18$

$x-y=2$ ……… $\left(1\right)$

Also given $x+y=6$ ……… $\left(2\right)$

Solving $\left(1\right)$ and $\left(2\right)$, we get

$x=4,y=2$

The required number is $10y+x=10\times 2+4=20+4=24$.