Question

The sum of the digits of a two-digit number is $12$. If the new number formed by reversing the digits is greater than the original number by $18$, find the original number.

Answer 1

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

Then the original number be $10x+y$.

Value of the number with reversed digits is $10y+x$.

As per question, we have

$x+y=12$ ....(1)

If the digits are reversed, the digits is greater than the original number by $18$.

Therefore, $10y+x=10x+y+18$

$\Rightarrow 9x-9y=-18$ ....(2)

Multiply equation (1) by $9$, we get

$9x+9y=108$ ....(3)

Add equations (2)and (3),

$18x=90$

$\Rightarrow x=5$

Substitute this value in equation (1), we get

$5+y=12\Rightarrow y=7$

Therefore, the original number is $10x+y=10\times 5+7=57$..