Knowing Letters

Consider first the units column. Since $c+f$ must end in $0$ ,then $c+f =0$ or $c+f =10$ .

The value of $c+f$ can not 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, $a + b + c + d + e + f = (a + d) + (b + e) + (c + f) = 9 + 9 + 10 = 28$ .

For $A =5, B = 2, C=5,D=4,E=7,$ and $F=5$

$\begin{array} { l l l l l } & & 5 & 2 & 5 \\+ && 4 &7 & 5 \\ \hline &\ 1 &0 &0 &0 \end{array}$

$A+B+C+D+E+F = 5+2+5+4+7+5 = \boxed{28}$

629 + 371 = 1000