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?

The correct option is B - 9.

[as per provisional answerkey]

Solution

Let the 2-digit number A be represented as $10x+y$, where x and y are digits (x ≠ 0 and x ≠ y).
Since B is obtained by reversing the digits of A, $B=10y+x$.

Given that $A>B$ and (A - B) is a multiple of 27.
$A-B=(10x+y)-(10y+x)$
$A-B=9x-9y$
$A-B=9(x-y)$

According to the question, $9(x-y)$ must be a multiple of 27.
Let $9(x-y)=27k$, where k is an integer ($k>0$ because $A>B$).
Dividing both sides by 9, we get:
$(x-y)=3k$

Since x and y are single digits (0 to 9), the possible values for the difference (x - y) that are multiples of 3 are 3, 6, and 9.

Case 1: $x-y=3$
Possible pairs (x, y) are: (4,1), (5,2), (6,3), (7,4), (8,5), (9,6).
(Note: x cannot be 3 because then y would be 0, and while A=30 is a 2-digit number, its reverse B=03 is often considered a 1-digit number in such problems. However, even if we include it, we must check the total count against the options.)
Pairs: (3,0), (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) - Total 7 pairs.

Case 2: $x-y=6$
Possible pairs (x, y) are: (6,0), (7,1), (8,2), (9,3) - Total 4 pairs.

Case 3: $x-y=9$
Possible pairs (x, y) are: (9,0) - Total 1 pair.

Total possible values for A = $7+4+1=12$.
Wait, let's re-examine the constraint "A is a 2-digit number with different digits" and the standard interpretation of B as a "2-digit number". If B must also be a 2-digit number, y cannot be 0.

If y ≠ 0:
Case 1 (diff 3): (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) - 6 values.
Case 2 (diff 6): (7,1), (8,2), (9,3) - 3 values.
Case 3 (diff 9): No values (since x would be 9 and y would be 0).
Total = $6+3=9$.

Why the other options are incorrect

• Option (a) - 6: This count only considers the case where the difference between digits is 3 and excludes the cases where the difference is 6 or 9.
• Option (c) - 12: This count includes cases where y = 0 (30, 60, 90). However, if A = 30, then B = 03, which is technically a 1-digit number, violating the condition that B is a "2-digit number".
• Option (d) - 18: This is an overestimation that might occur if one incorrectly assumes the difference (x-y) could be any multiple of 3 and fails to restrict the digits to the 1-9 range for both x and y.

Key Concept

The difference between a two-digit number and its reverse is always 9 times the difference of its digits; for this to be a multiple of 27, the difference between the digits must be a multiple of 3.