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?
Answer & explanation
Answer: (b) 9
A − B = 9 × (difference of digits), so the digit difference must be a multiple of 3: 3 or 6 (9 would need a 0 as the units digit, making B a one-digit number). That gives 6 + 3 = 9 numbers.
- A = 10a + b, B = 10b + a, so A − B = 9(a − b), with a > b.
- 9(a − b) is a multiple of 27 when a − b is 3, 6 or 9.
- B must be two-digit, so b ≥ 1. a − b = 9 would need b = 0 — not allowed.
- a − b = 3: 41, 52, 63, 74, 85, 96 → 6 numbers.
- a − b = 6: 71, 82, 93 → 3 numbers.
- Total = 9. (Allowing b = 0 would give 12 — the trap option.)
Remember · A two-digit number minus its reverse = 9 × (difference of the digits). Remember that the reverse must stay two-digit.
Question and answer: UPSC's provisional GS Paper II (2026, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·