A 2-digit number is reversed. The larger of the two numbers is divided by the smaller one. What is the largest possible remainder?
Answer & explanation
Answer: (d) 45
A number and its reverse differ by 9 × (difference of the digits). When the larger is less than twice the smaller, the quotient is 1 and the remainder is that difference; the biggest such case is 94 ÷ 49, which leaves 45.
- Let the larger number be 10a + b and the smaller 10b + a, with a > b.
- Their difference is 9(a − b). If the larger is less than twice the smaller, the quotient is 1 and the remainder is exactly 9(a − b).
- Quotient 1 needs 10a + b < 2(10b + a), i.e. 8a < 19b. With a = 9 this needs b ≥ 4, so the best pair is 94 and 49: remainder 9 × 5 = 45.
- If the quotient is 2 or more, then 8a ≥ 19b forces b ≤ 3, so the smaller number is at most 39 and the remainder is less than 39.
- Largest possible remainder = 45 (94 = 1 × 49 + 45).
Remember · A number and its reverse differ by 9 × (difference of digits); when the quotient is 1, that difference is the remainder.
Question and answer: UPSC's official GS Paper II (2017, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·