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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 Problem, Question \#11) 
Overseas Competition->Knowledge Point->Counting Modules
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 $(\text{B}) 7$ ordered pairs. 