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 ?

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Accepted Answer

The correct answer is option D. We are given ABC + DEF = 1111, where A, B, C, D, E, F are distinct non-zero digits. Let's analyze the addition column by column:
1. Units place: C + F must result in a unit digit of 1. Since C and F are distinct non-zero digits, C + F = 11 (with a carry of 1 to the tens place).
2. Tens place: B + E + (carry 1) = 11. So, B + E = 10 (with a carry of 1 to the hundreds place).
3. Hundreds place: A + D + (carry 1) = 11. So, A + D = 10.
We need to find distinct non-zero digits for A, B, C, D, E, F that satisfy these conditions. One possible set of assignments is:
- C + F = 11: Let C=2, F=9.
- B + E = 10: Using remaining digits, let B=3, E=7.
- A + D = 10: Using remaining digits, let A=4, D=6.
All digits (2, 9, 3, 7, 4, 6) are distinct and non-zero. Let's verify: 432 + 679 = 1111. This works. The sum A+B+C+D+E+F = 4+3+2+6+7+9 = 31. Any valid combination of distinct non-zero digits satisfying the sums will yield the same total sum. This question tests logical deduction and number properties, a typical CSAT problem.

Q39. 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 ?

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