The sum of digits of a …

Let the digits at units and tens place of the given number be x and y respectively.
Thus, the number is {tex}10y + x{/tex}.
The sum of the digits of the number is 13.
Thus, we have {tex}x + y = 13{/tex}

According to the question,
After interchanging the digits, the number becomes {tex}10x + y{/tex}.
The difference between the number obtained by interchanging the digits and the original number is 45.
Thus, we have {tex}(10x + y) - (10y + x) = 45{/tex}
{tex}\Rightarrow{/tex} {tex}10x + y - 10y - x = 45{/tex}
{tex}\Rightarrow{/tex} {tex}9x - 9y =45{/tex}
{tex}\Rightarrow{/tex} {tex}9(x - y) = 45{/tex}
{tex}\Rightarrow{/tex} {tex}x - y = 5{/tex}
So, we have two equations
{tex}x + y = 13{/tex}
{tex}x - y = 5{/tex}
Here x and y are unknowns.
We have to solve the above equations for x and y.
Adding the two equations, we have
{tex}(x + y) + (x - y) = 13 + 5{/tex}
{tex}\Rightarrow{/tex} {tex}x + y + x - y =18{/tex}
{tex}\Rightarrow{/tex} {tex}2x = 18{/tex}
{tex}\Rightarrow{/tex} {tex}x = 9{/tex}
Substituting the value of x in the first equation, 
{tex}\Rightarrow{/tex} {tex}9 + y = 13{/tex}
{tex}\Rightarrow{/tex} {tex}y = 13 - 9{/tex}
{tex}\Rightarrow{/tex} {tex}y = 4{/tex}
Hence, the number is 10 {tex}\times{/tex} 4 + 9 = 49