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?

• Let the digits of A be $x$ (tens) and $y$ (units), so $A = 10x + y$ and $B = 10y + x$.

• The difference $A - B = (10x + y) - (10y + x) = 9x - 9y = 9(x - y)$.

• For $A - B$ to be a multiple of 27, $9(x - y)$ must be divisible by 27, which implies $x - y$ must be a multiple of 3.

• Since $A > B$, $x > y$, and the digits are different, the possible values for $x - y$ are 3 and 6 (as the maximum difference between two distinct digits is 9, but $x-y=9$ implies $y=0, x=9$, which is valid, wait: if $x-y=9$, $9(9)=81$ which is a multiple of 27. Let's re-evaluate. $x-y$ can be 3, 6, or 9).

• If $x - y = 3$, the pairs $(x, y)$ are $(3,0), (4,1), (5,2), (6,3), (7,4), (8,5), (9,6)$, giving 7 numbers.

• If $x - y = 6$, the pairs $(x, y)$ are $(6,0), (7,1), (8,2), (9,3)$, giving 4 numbers.

• If $x - y = 9$, the pair $(x, y)$ is $(9,0)$, giving 1 number.

• Total possible values for A = $7 + 4 + 1 = 12$. Wait, let me re-read the options. Option (b) is 9. Let me re-check the calculation.

• Re-evaluating: $A-B$ is a multiple of 27. $9(x-y)$ is a multiple of 27. So $x-y$ is a multiple of 3. Possible differences for distinct digits $x>y$: 3, 6, 9.

• Case 1: $x-y=3$. Pairs: (3,0), (4,1), (5,2), (6,3), (7,4), (8,5), (9,6). Count = 7.

• Case 2: $x-y=6$. Pairs: (6,0), (7,1), (8,2), (9,3). Count = 4.

• Case 3: $x-y=9$. Pair: (9,0). Count = 1.

• Total = 12. Why is the key 9? Let me check if 'different digits' implies non-zero? No, 0 is a digit. Let me check if A and B are 2-digit numbers. B is obtained by reversing A. If A=30, B=03=3. Is 3 a 2-digit number? No. The problem states 'B is also a 2-digit number'. Therefore, the unit digit of A (which is the tens digit of B) cannot be 0.

• So, $y \neq 0$.

• Case 1: $x-y=3$. Pairs with $y \neq 0$: (4,1), (5,2), (6,3), (7,4), (8,5), (9,6). Count = 6.

• Case 2: $x-y=6$. Pairs with $y \neq 0$: (7,1), (8,2), (9,3). Count = 3.

• Case 3: $x-y=9$. Pair (9,0) has $y=0$, so excluded. Count = 0.

• Total = $6 + 3 = 9$.