$A$ 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?

Question

$A$ 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?

Accepted Answer

$9$

Explanation: Let A = 10a + b and B = 10b + a, where a and b are digits with a ≠ b and b ≠ 0 (so B remains 2-digit). A − B = 9(a − b). For this to be a multiple of 27, we need 9(a − b) divisible by 27, so (a − b) must be a multiple of 3. Since A > B, we have a > b, giving possible differences: 3, 6, or 9. For a − b = 3: pairs (4,1), (5,2), (6,3), (7,4), (8,5), (9,6) = 6 numbers. For a − b = 6: pairs (7,1), (8,2), (9,3) = 3 numbers. For a − b = 9: pair (9,0) is invalid since B would not be 2-digit. Total = 6 + 3 = 9. Therefore, 9 different values of A are possible, making Option 2 the correct answer.

Detailed Explanation:

Question 6 of 7 HCF and LCM

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