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?

1. S is always divisible by 74.

2. S is always divisible by 9.

Select the correct answer using the code given below:

The correct answer is Both 1 and 2.

Step 1: Identify the eligible digits

The question specifies non-zero digits that are multiples of 3.
The single-digit non-zero multiples of 3 are 3, 6, and 9.
Since we need to form 3-digit numbers using three non-zero digits (without repetition of digits), the three digits used must be 3, 6, and 9.

Step 2: Form all possible 3-digit numbers

Using the digits 3, 6, and 9 without repetition, we can form the following 3-digit numbers:

369, 396, 639, 693, 936, and 963.

Step 3: Calculate the sum S

Let us find the sum S of all these numbers:

$S=369+396+639+693+936+963$

Summing them up step-by-step:

$369+396=765$

$765+639=1404$

$1404+693=2097$

$2097+936=3033$

$3033+963=3996$

So, the sum S = 3996.

General Algebraic Approach:
For any three digits a, b, and c, the sum of all 6 possible 3-digit permutations is given by:

$S=222\×(a+b+c)$

Here, a = 3, b = 6, c = 9.

$a+b+c=3+6+9=18$

$S=222\×18=3996$

Step 4: Test Statement 1 (Divisibility by 74)

Let us check if 3996 is divisible by 74:

$\frac{3996}{74}=54$

Since 3996 divided by 74 gives an exact integer quotient of 54, S is divisible by 74.
Thus, Statement 1 is correct.

Step 5: Test Statement 2 (Divisibility by 9)

Let us check if 3996 is divisible by 9:
Sum of the digits of 3996 is 3 + 9 + 9 + 6 = 27, which is a multiple of 9.

$\frac{3996}{9}=444$

Since 3996 divided by 9 gives an exact integer quotient of 444, S is divisible by 9.
Thus, Statement 2 is also correct.

Conclusion:

Both statements 1 and 2 are correct. Hence, the correct code option is Both 1 and 2.