The digits 1 to 9 are arranged in three rows in such a way that each row contains three digits, and the number formed in the second row is twice the number formed in the first row; and the number formed in the third row is thrice the number formed in the first row. Repetition of digits is not allowed. If only three of the four digits 2, 3, 7 and 9 are allowed to use in the first row, how many such combinations are possible to be arranged in the three rows?
Answer & explanation
Answer: (c) 2
The first-row number N must keep 3N below 1000, so it starts with 2 or 3 and uses three of 2, 3, 7, 9. Testing the eight such numbers leaves only 273 (546, 819) and 327 (654, 981).
- The third row is 3N, a 3-digit number, so N ≤ 333. With digits from 2, 3, 7, 9 (no repeats), N must start with 2 or 3.
- Candidates: 237, 239, 273, 279, 293, 297, 327 and 329.
- 237 → 474 (repeats 4); 239 → 478, 717 (repeats 7); 279 → 558 (repeats 5); 293 → 586, 879 (repeats 8 and 9); 297 → 594 (repeats 9); 329 → 658, 987 (repeats 8).
- 273 → 546 → 819 uses each of 1–9 once; 327 → 654 → 981 also uses each of 1–9 once.
- So 2 arrangements are possible.
- Check: 273 × 2 = 546 and 273 × 3 = 819; 327 × 2 = 654 and 327 × 3 = 981.
Remember · Bound the search first (here 3N < 1000, so N ≤ 333); a short, systematic list beats guessing.
Question and answer: UPSC's official GS Paper II (2022, Series A) — paper ↗ · answer key ↗. Explanation: Minimalist IAS, checked 30 Sept 2026 (how we verify). ·