Consider all 3-digit numbers (without repetition of digits) obtained using three non-zero digits which are multiples of 3. Let S be their sum. 1. S is always divisible by 74. 2. S is always divisible by 9. Which of the following is/are correct?

1. **Identify the Digits**: The non-zero digits which are multiples of 3 are {3, 6, 9}.\n\n2. **Structure the Sum**: From analyzing previous year questions [1], we know that the sum of numbers formed by permuting digits follows a pattern. For 3 distinct digits, the sum of all 6 possible numbers is: \n Sum = (Sum of digits) x 222.\n\n3. **Analyze Statement 1 (Divisibility by 74)**: \n - We look at the multiplier 222. \n - Breaking it down: 222 = 2 x 111.\n - From our study of 'repunts' [8] [11], we know 111 = 3 x 37.\n - So, 222 = 2 x 3 x 37 = 3 x (2 x 37) = 3 x 74.\n - Since 222 is a multiple of 74, the total Sum is *always* divisible by 74, regardless of which digits are used. \n - **Statement 1 is Correct**.\n\n4. **Analyze Statement 2 (Divisibility by 9)**:\n - We check the sum of the digits: 3 + 6 + 9 = 18.\n - The total Sum = 18 x 222.\n - Since 18 is divisible by 9, the entire result is divisible by 9 [73].\n - **Statement 2 is Correct**.\n\n5. **Conclusion**: Both statements are independently true.