We can write the two digit number in the form of $10 a+b$; reverse of $10 a+b$ is $10 b+a$. The sum of those numbers is:  $$ \begin{gathered} (10 a+b)+(10 b+a)=132 \textbackslash\textbackslash{} 11 a+11 b=132 \textbackslash\textbackslash{} a+b=12 \end{gathered} $$  We can use brute force to find order pairs $(a, b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$.  Thus our ordered pairs are $(3,9) ;(4,8) ;(5,7) ;(6,6) ;(7,5) ;(8,4) ;(9,3)$ or $(\mathbf{B}) 7$ ordered pairs. 
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132$. (2016 AMC 8 Problems, Question \#11) 
- $$5$$ 
- $$7$$ 
- $$9$$ 
- $$11$$ 
- $$12$$ 