A 3-digit number ABC, on multiplication with D gives 37DD where A, B, C and D are different non-zero digits. What is the value of A + B + C?
Answer & explanation
Answer: (a) 18
37DD = 3700 + 11D must be a multiple of D, so D must divide 3700; that allows D = 1, 2, 4 or 5. Only D = 4 gives a 3-digit ABC with all four digits different: 3744 ÷ 4 = 936, so A + B + C = 9 + 3 + 6 = 18.
- 37DD = 3700 + 11 × D. Since ABC × D = 37DD, D divides 3700 + 11D, so D divides 3700 = 2² × 5² × 37.
- Possible non-zero digits: D = 1, 2, 4 or 5.
- D = 1 gives ABC = 3711 and D = 2 gives 3722 ÷ 2 = 1861: neither is a 3-digit number.
- D = 5 gives 3755 ÷ 5 = 751, but then B = 5 = D, which is not allowed.
- D = 4 gives 3744 ÷ 4 = 936: A = 9, B = 3, C = 6, D = 4, all different. A + B + C = 18.
- Check: 936 × 4 = 3744.
Remember · In digit puzzles, write the unknown number algebraically (3700 + 11D), shortlist digits by divisibility, then apply the 'different digits' condition.
Question and answer: UPSC's official GS Paper II (2023, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·