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 54, find the original number. Check your solution.

Answer 1

Let the units digit of the two-digit number be $y$ and the tens digit be $x$.

Hence, the number will be $10x+y$.

Given that the sum of digits is $12$.

Hence, $x+y=12.......\left(i\right)$

It is also given that the new number formed by reversing the digits is greater than the original number by $54$.

On reversing the digits, $x$ becomes the units digit and $y$ becomes the tens digit.

Hence, the new number will be $10y+x$.

$\therefore 10y+x=10x+y+54$

$\Rightarrow 9y-9x=54$

$\Rightarrow y-x=6$

$\Rightarrow -x+y=6.........\left(ii\right)$

Adding $\left(i\right)$ and $\left(ii\right)$, we get:

$(x+y)+(-x+y)=12+6$

$\Rightarrow 2y=18$

$\therefore y=\frac{18}{2}=9$

Substituting $y=9$ in $\left(i\right)$, we get:

$x+9=12$

$\therefore x=12-9=3$

The units digit is $3$ and the tens digit is $9$.

Hence, the number is $39$.

Checking the solution:

Original number $=39$

Hence, number obtained by reversing the digits $=93$

$93=39+54$

Hence, the number obtained by reversing the digits is greater than the original number by $54$.

Hence, the answer is correct.