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 ?
- A28
- B29
- C30
- D31Correct
Explanation
To solve this, write the addition in vertical form: A B C
- D E F
1 1 1 1
Looking at the units place, C + F must end in 1. Since A, B, C, D, E, and F are non-zero digits, the maximum sum of two digits is 9 + 8 = 17. Therefore, C + F must be 11. This results in a carryover of 1 to the tens place.
In the tens place, we have the carryover 1 + B + E. This sum must also end in 1. For this to happen, 1 + B + E must be 11, which means B + E = 10. This results in another carryover of 1 to the hundreds place.
In the hundreds place, we have the carryover 1 + A + D. This sum must equal 11 to complete the number 1111. Therefore, 1 + A + D = 11, which means A + D = 10.
Now, add the pairs together: (A + D) + (B + E) + (C + F) = 10 + 10 + 11 = 31.
Thus, the total sum of A + B + C + D + E + F is 31. The correct option is D.
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