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.40
- B.3
- C.2
- D.1
▶ Answer & Explanation
Correct answer: C. 2
Let the number in the first row be X. The numbers in the second and third rows are 2X and 3X respectively. The digits used for X, 2X, and 3X must be a permutation of the digits 1 through 9, with no repetitions. This implies that X must be a 3-digit number. The total sum of digits 1-9 is 45. The sum of digits in X, 2X, and 3X must also be a permutation of 1-9, so their sum is 45. This means the sum of the digits of X, 2X, and 3X must also be divisible by 9. A systematic check of possible values for X, keeping in mind the constraints (digits 1-9, no repetition, 2X and 3X also using distinct digits from 1-9), reveals only a few solutions. The constraint that only three out of {2, 3, 7, 9} can be used in the first row (X) further limits the possibilities.
Source: UPSC csat 2022