the sum of the digits in the number is 13. the number is 27 more than the number formed by interchanging the tens digit with the one digit. What is the number?
I'm guessing: You are looking for a two-digit-number(?).
If so:
Let x, y denote the digits with $\displaystyle x\in\{1, 2, ..., 9\}~\wedge~y\in\{0,1,2,...,9\}$
The original number is 10x + y
and the number with reverse order of digits is 10y + x
You have a system of simultaneous equations:
$\displaystyle \left|\begin{array}{rcl}x+y&=&13\\10x+y&=&10y+x+27 \end{array}\right.$
Solve for x and y. I've got x = 8 and y = 5