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.
Which of the following is/are correct?

Correct Option (3)

The non-zero digits that are multiples of 3 are 3, 6, and 9.

Using these three distinct digits, the 3-digit numbers that can be formed without repetition are: 369, 396, 639, 693, 936, and 963.

The sum (S) of all such numbers can be calculated using the formula:

S = (Sum of the digits) × (Number of permutations of the remaining digits) × (Sum of place values for a number of 'n' digits)

For 3 distinct digits (3, 6, 9):

• Sum of digits = 3 + 6 + 9 = 18
• Number of permutations of the remaining (3-1) = 2 digits = 2! = 2
• Sum of place values for a 3-digit number = 100 + 10 + 1 = 111

Therefore, S = 18 × 2 × 111 = 36 × 111 = 3996.

Statement 1: S is always divisible by 74.

3996 ÷ 74 = 54.

Since 3996 is perfectly divisible by 74, Statement 1 is correct.

Statement 2: S is always divisible by 9.

The sum of the digits of S (3996) is 3 + 9 + 9 + 6 = 27.

Since 27 is divisible by 9, S is also divisible by 9. Thus, Statement 2 is correct.

As both Statement 1 and Statement 2 are correct, option (3) is the correct answer.

Incorrect Options:

Options (1), (2), and (4) are incorrect. The analysis confirms that both Statement 1 and Statement 2 are factually correct regarding the divisibility of the sum S by 74 and 9, respectively. Therefore, selecting only one statement as correct or neither statement as correct would be erroneous.