Two 3-digit numbers, $a b c$ and def, have the following property:

| $a b c$ |
| ---: |
| $+\quad d \quad e f$ |
| 1000 |

None of $a, b, c, d, e$, or $f$ is 0 . What is $a+b+c+d+e+f$ ?
(A) 10
(B) 19
(C) 21
(D) 28
(E) 30
Solution 1

The wording of the problem tells us that $a+b+c+d+e+f$ must be the same no matter what numbers $a b c$ and $d e f$ are chosen that satisfy the conditions.

An example that works is $889+111=1000$.

In this case, $a+b+c+d+e+f=8+8+9+1+1+1=28$, so this must always be the value.

## Solution 2

Consider performing this "long addition" by hand.

Consider first the units column.

Since $c+f$ ends in a 0 , then $c+f=0$ or $c+f=10$. The value of $c+f$ cannot be 20 or more, as $c$ and $f$ are digits.

Since none of the digits is 0 , we cannot have $c+f=0+0$ so $c+f=10$. (This means that we "carry" a 1 to the tens column.)

Since the result in the tens column is 0 and there is a 1 carried into this column, then $b+e$ ends in a 9 , so we must have $b+e=9$. (Since $b$ and $e$ are digits, $b+e$ cannot be 19 or more.) In the tens column, we thus have $b+e=9$ plus the carry of 1 , so the resulting digit in the tens column is 0 , with a 1 carried to the hundreds column.

Using a similar analysis in the hundreds column to that in the tens column, we must have $a+d=9$.

Therefore, $a+b+c+d+e+f=(a+d)+(b+e)+(c+f)=9+9+10=28$.

ANSWER: (D)