If ABC and DEF are both 3-digit numbers such that A, B, C, D, E, and F are distinct non-zero digits such that ABC+ DEF= 1111, then what is the value of A+B+C+D+E+F?

Here's a brief explanation: 1. **Break down the sum by place value:** When we add two 3-digit numbers ABC and DEF to get 1111, we can analyze it column by column, from right to left (units, tens, hundreds). 2. **Units place:** C + F must result in a unit digit of 1. Since C and F are distinct non-zero digits (minimum sum 1+2=3), C+F cannot be 1. Therefore, C + F = 11. This generates a carry-over of 1 to the tens place. 3. **Tens place:** B + E + (carry-over from units) must result in a unit digit of 1. So, B + E + 1 = 11. This implies B + E = 10. This generates a carry-over of 1 to the hundreds place. 4. **Hundreds place:** A + D + (carry-over from tens) must result in a unit digit of 1. So, A + D + 1 = 11. This implies A + D = 10. This generates a carry-over of 1 to the thousands place. 5. **Thousands place:** The carry-over of 1 from the hundreds place matches the thousands digit in 1111. 6. **Calculate the total sum:** We need to find A+B+C+D+E+F. This can be written as (A+D) + (B+E) + (C+F). Substituting the sums we found: 10 + 10 + 11 = 31. 7. **Verify distinct non-zero digits (optional, but good for confidence):** We can find such digits. For example: C=2, F=9 (C+F=11) B=3, E=7 (B+E=10) A=4, D=6 (A+D=10) All digits {2,9,3,7,4,6} are distinct and non-zero. ABC = 432, DEF = 679. 432 + 679 = 1111. The sum of these digits is 4+3+2+6+7+9 = 31. The value of A+B+C+D+E+F is 31. The final answer is D) 31.