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 $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$
Solution 1 - kindlymath55532
We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is:
\[(10a+b)+(10b+a)=132\]\[11a+11b=132\]\[a+b=12\]
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 $\boxed{\textbf{(B)} 7}$ ordered pairs.

Solution 2 - SweetMango77
Since the numbers are “mirror images,” their average has to be $\frac{132}{2}=66$. The highest possible value for the tens digit is $9$ because it is a two-digit number. $9-6=3$ and $6-3=3$, so our lowest tens digit is $3$. The numbers between $9$ and $3$ inclusive is $9-3+1=\boxed{\text{(B)}\;7}$ total possibilities.