A 4-digit number N is such that when divided by 3, 5, 6, 9 leaves a remainder 1, 3, 4, 7 respectively. What is the smallest value of N?
Answer & explanation
Answer: (c) 1078
Each remainder is 2 less than its divisor, so N + 2 is a multiple of LCM(3, 5, 6, 9) = 90. The smallest 4-digit number of the form 90k − 2 is 1080 − 2 = 1078.
- In every case the remainder is 2 short of the divisor: 3 − 1 = 5 − 3 = 6 − 4 = 9 − 7 = 2.
- So N + 2 is divisible by 3, 5, 6 and 9, i.e. by their LCM, 90.
- N = 90k − 2. k = 11 gives 988 (only 3 digits); k = 12 gives 1080 − 2 = 1078.
- Check: 1078 = 3 × 359 + 1 = 5 × 215 + 3 = 6 × 179 + 4 = 9 × 119 + 7.
Remember · If divisor minus remainder is the same d for every divisor, the number is (LCM × k) − d.
Question and answer: UPSC's official GS Paper II (2025, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·