The sum of the digits of a two-digit number is $13.$ The difference between the number and the number with its digits reversed is $27.$ What is the sum of the original number and the number with its digits reversed?
Let the two-digit number be $10x+y$, where $x$ and $y$ are digits from $0$ to $9$. Then the number with its digits reversed is $10y+x$. The sum of the digits is $13$, so $x+y=13$. The difference between the number and the number with its digits reversed is $27$, so $10x+y-(10y+x)=27$. Simplifying, we get $9x-9y=27$, or $x-y=3$. Solving the system of equations $x+y=13$ and $x-y=3$, we get $x=8$ and $y=5$. So the original number is $85$ and the number with its digits reversed is $58$. The sum of the original number and the number with its digits reversed is $85+58=143$. 