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?
- (a) 6
- (b) 9 ✓ UPSC's answer
- (c) 12
- (d) 18
Why the answer is (b)
• Write A = 10a + b and B = 10b + a, so A − B = 9(a − b).
• 9(a − b) is a multiple of 27 when a − b is a multiple of 3. With a > b, a − b = 3, 6 or 9.
• a − b = 9 means A = 90 and B = 09, which is not a two-digit number, so it is ruled out. B two-digit also needs b ≥ 1.
• a − b = 3: 41, 52, 63, 74, 85, 96 (6 numbers). a − b = 6: 71, 82, 93 (3 numbers). Total 9: (b).
Why the other options are wrong
- (a) 6
- 6 counts only the a − b = 3 cases and misses 71, 82, 93.
- (c) 12
- 12 includes numbers whose reverse is not a two-digit number.
- (d) 18
- 18 does not respect A > B and two-digit B.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2026, held on 24 May 2026. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.