jmc

The two digit number can be represented as $10x+y,$ where $x$ and $y$ are digits, with $x\ne 0.$ We are given that the sum of the digits is $13,$ so $x+y=13.$ If we reverse the digits of this number, we have $10y+x.$ We are given that the difference is $27,$ but we don't know if the original number or if the number with its digits reversed is greater. We can show this as such: $$|(10x+y)-(10y+x)|=27.$$ However, it doesn't matter which of the two numbers is greater, since we wish to find their sum. So, without loss of generality, we will let the first number be the larger of the two. This means that $x>y,$ so we can get rid of the absolute values in our last equation to obtain $9x-9y=27,$ equivalent to $x-y=3.$

We now have two equations in two variables: $x+y=13$ and $x-y=3.$ Adding the two, we obtain $2x=16,$ so $x=8.$ Subtracting, we obtain $2y=10,$ so $y=5.$ Thus, the original number is $85,$ and our answer is $85+58=\boxed{143}.$

OR

As before, the two digit number can be expressed as $10x+y,$ and the number with its digits reversed is $10y+x.$ We want to find the sum of these two numbers, which is $$(10x+y)+(10y+x)=11x+11y=11(x+y).$$ We are given that the sum of the digits is $13,$ so $x+y=13.$ Since all we want is $11(x+y),$ we can substitute for $x+y$ to obtain our answer of $11\cdot 13=\boxed{143}.$