Correct Option (A)

The problem states that A, B, C, and D are different non-zero digits. A 3-digit number ABC, when multiplied by D, yields the 4-digit number 37DD.

The multiplication can be represented as: (100A + 10B + C) × D = 3700 + 10D + D = 3700 + 11D.

Let's analyze the multiplication based on place values:

Units Place: The product C × D must result in a number whose unit digit is D.
- Possible (C, D) pairs (where C, D are different non-zero digits and C × D ends in D): (3,5), (6,4), (7,5), (9,5).
- From the product 37DD, the tens digit is D and the units digit is D. This implies D is a specific single digit.
- Considering the overall product is 37DD, the thousands digit is 3 and the hundreds digit is 7.
- If we consider the structure 37DD, the tens digit is D. This implies that the result of the tens place calculation (B × D + carry-over from C × D) must have D as its unit digit.
- Let's deduce D first. If D were 5 (from pairs (3,5), (7,5), (9,5)), then ABC × 5 = 3755. This would mean 3755 is divisible by 5, which is true. However, the last digit of 37DD is D, so if D=5, the number is 3755.
- If D were 4 (from pair (6,4)), then ABC × 4 = 3744. This means 3744 is divisible by 4, which is true. The last digit of 37DD is D, so if D=4, the number is 3744.
- Let's test D=4. If D=4, then C × 4 must end in 4. Possible C values (non-zero, different from D=4) are C=1 (1 × 4 = 4) or C=6 (6 × 4 = 24). Since A, B, C, D must be different, C cannot be 1 (as D=4). Therefore, C=6 is the only possibility for D=4.
  - So, C=6 and D=4. C × D = 6 × 4 = 24. The unit digit is 4 (D), and the carry-over to the tens place is 2.
Tens Place: The tens digit of the product 37DD is D (which is 4). The calculation involves (B × D) plus the carry-over from the units place.
- Using D=4 and the carry-over of 2 from C × D: (B × 4) + 2 must result in a number whose unit digit is 4.
- This implies B × 4 must end in 2.
- Possible non-zero digits for B, distinct from C=6 and D=4, are 3 (since 3 × 4 = 12) or 8 (since 8 × 4 = 32).
- If B=3: (3 × 4) + 2 = 12 + 2 = 14. The unit digit is 4 (D), and the carry-over to the hundreds place is 1.
- If B=8: (8 × 4) + 2 = 32 + 2 = 34. The unit digit is 4 (D), and the carry-over to the hundreds place is 3.
Hundreds and Thousands Place: The first two digits of the product 37DD are 37. The calculation involves (A × D) plus the carry-over from the tens place.
- Case 1: Assume B=3, with a carry-over of 1 to the hundreds place.
  - (A × 4) + 1 = 37.
  - A × 4 = 36.
  - A = 9.
  - This gives A=9, B=3, C=6, D=4. All are distinct non-zero digits (9, 3, 6, 4).
- Case 2: Assume B=8, with a carry-over of 3 to the hundreds place.
  - (A × 4) + 3 = 37.
  - A × 4 = 34.
  - A = 34/4 = 8.5, which is not an integer. Therefore, B cannot be 8.

Thus, the unique set of digits is A=9, B=3, C=6, D=4.

Verification: 936 × 4 = 3744. This matches the form 37DD, where D=4.

The value of A + B + C = 9 + 3 + 6 = 18.

Incorrect Options:

Options 2 (16), 3 (15), and 4 (Cannot be determined due to insufficient data) are incorrect because a unique solution for A, B, and C can be derived from the given conditions, leading to A + B + C = 18.