The sum of a two digit number and the number obtained by reversing the order of its digits is 99. If the digits differ by 3, find the number.
Let the digits at units and tens place of the given number be $x$ and $y$ respectively. Thus, the number is $10 y+x$.
The two digits of the number are differing by 3 . Thus, we have $x-y=\pm 3$
After interchanging the digits, the number becomes $10 x+y$.
The sum of the numbers obtained by interchanging the digits and the original number is 99 . Thus, we have
$(10 x+y)+(10 y+x)=99$
$\Rightarrow 10 x+y+10 y+x=99$
$\Rightarrow 11 x+11 y=99$
So, we have two systems of simultaneous equations
Hence, the number is $10 \times 3+6=36$.
(ii) Now, we solve the system
Adding the two equations, we have
$\Rightarrow 2 x=6$
Hence, the number is $10 \times 6+3=63$.
Note that there are two such numbers.