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?
- (a) 4
- (b) 3
- (c) 2 ✓ UPSC's answer
- (d) 1
Why the answer is (c)
• Let the first-row number be N; the other rows are 2N and 3N, and together the nine digits must be 1 to 9 with no repetition.
• 3N must have three digits, so N ≤ 333; hence N starts with 2 or 3 and uses three of the digits 2, 3, 7, 9.
• The candidates are 237, 239, 273, 279, 293, 297, 327 and 329.
• 237 → 474 (repeated 4); 239 → 478, 717 (repeated 7); 279 → 558 (repeated 5); 293 → 586, 879 (8 and 9 repeated); 297 → 594 (9 repeated); 329 → 658, 987 (8 and 9 repeated).
• 273 → 546 → 819 uses every digit 1–9 exactly once, and so does 327 → 654 → 981.
• So exactly two arrangements are possible, option (c).
Why the other options are wrong
- (a) 4
- Only 273 and 327 work, so there are 2 arrangements, not 4.
- (b) 3
- Only 273 and 327 work, so there are 2 arrangements, not 3.
- (d) 1
- Both 273 (546, 819) and 327 (654, 981) work, so there are 2 arrangements, not 1.
Asked in the CSAT Paper II of the UPSC Civil Services Preliminary Examination 2022, held on 5 June 2022. Question and answer key: Union Public Service Commission. Explanation: UPSC Answer Check.