A is a 2-digit number with different digits. B is also a 2-digit number and is obtained by reversing the digits of A. If A - B is a multiple of 27, where A > B, how many such different A's are possible?

Correct Answer: (B) 9

Let $A = 10u + v$ and $B = 10v + u$ with non-zero digits $u, v$$\in$ \{1, $\dots$, 9\} and $u > v$. The difference is $A - B = 9(u - v)$. For $A - B$ to be a positive multiple of 27, $u - v$ must be a multiple of 3, so $u - v$$\in$ \{3, 6\} (since digits differ by at most 8). For $u - v = 3$, the valid pairs $(u, v)$ are (4,1), (5,2), (6,3), (7,4), (8,5), and (9,6) (6 values of A). For $u - v = 6$, the valid pairs are (7,1), (8,2), and (9,3) (3 values of A). Thus, there are $6 + 3 = 9$ such different values for A.