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
This puzzle involves arranging digits 1-9 into three 3-digit numbers (let's call them N1, N2, N3) such that N2 = 2 * N1 and N3 = 3 * N1. All nine digits must be unique. The constraint is that the digits used in N1 must be a subset of {2, 3, 7, 9}, with exactly three digits from this set forming N1. We test possible values for N1. If N1=132, N2=264, N3=396; all digits are unique. However, N1 does not use digits from the allowed set. If N1=237, N2=474 (repetition), N3=711 (repetition). If N1=273, N2=546, N3=819. The digits in N1 are 2, 7, 3, which are from the allowed set. The digits across all three numbers are {2, 7, 3, 5, 4, 6, 8, 1, 9}, which are unique and cover 1-9. This combination is valid. If N1=297, N2=594, N3=891. Digits in N1 are 2, 9, 7. Digits across all three are {2, 9, 7, 5, 9, 4, 8, 9, 1} which has repeated 9s and 7s. Invalid. If N1=327, N2=654, N3=981. Digits in N1 are 3, 2, 7. Digits across all three are {3, 2, 7, 6, 5, 4, 9, 8, 1}, which are unique and cover 1-9. This is valid. If N1=372, N2=744 (repetition). Invalid. If N1=723, N2=1446 (4 digits). Invalid. If N1=732, N2=1464 (4 digits). Invalid. If N1=927, N2=1854 (4 digits). Invalid. If N1=972, N2=1944 (4 digits). Invalid. Thus, only two combinations are possible using the allowed digits for N1.
Source: UPSC csat 2022