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 Option: (b) 9

Let the 2-digit number A be represented as 10x + y, where x is the tens digit and y is the units digit (with x not equal to y and x, y in {1..9}). The number B obtained by reversing the digits is 10y + x. Given that A > B, which means x > y. The difference A - B = (10x + y) - (10y + x) = 9(x - y). We are given that A - B is a multiple of 27, so 9(x - y) is a multiple of 27, which implies (x - y) must be a multiple of 3. Since x and y are digits (0 to 9) and x > y, the possible integer values for the difference (x - y) can be 3, 6, or 9. Let's count the number of valid pairs (x, y) for each difference: 1. If x - y = 3: possible pairs (x, y) are (3,0), (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) -> 7 pairs. 2. If x - y = 6: possible pairs (x, y) are (6,0), (7,1), (8,2), (9,3) -> 4 pairs. 3. If x - y = 9: possible pairs (x, y) are (9,0) -> 1 pair. Total number of different numbers A = 7 + 4 + 1 = 12? Wait, let's re-read carefully: B is a 2-digit number, which means the tens digit of B (which is y) cannot be 0! If y = 0, then B becomes a 1-digit number (e.g., if A = 30, B = 03 = 3, which is not a 2-digit number). Therefore, y must be greater than or equal to 1. Let's exclude cases where y = 0: 1. For x - y = 3: pairs are (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) -> 6 pairs. 2. For x - y = 6: pairs are (7,1), (8,2), (9,3) -> 3 pairs. 3. For x - y = 9: no pair with y > 0 (since if y=0, x=9, y is 0). Total pairs = 6 + 3 = 9 pairs. Thus, there are 9 such different A's possible. Option (b) is correct.