Letter Sum #1

    A A A C C E

+   F B G H D H

  H E C H D F A

    3 3 3 4 4 2

+   9 0 8 1 5 1

  1 2 4 1 5 9 3

Notes on solution, by puzzle editor:

For the purpose of this discussion, the 1st column is the rightmost column.

When adding two numbers, the carry is never more than 1, so you can immediately substitute H=1. And since no number is written with leading zeros, you know that neither A nor F is 0. Therefore, based on the 1st column, E+1=A; A is not 0 so E must be less than 9 and there is no carry to the 2nd column. And either A+F=10+E or 1+A+F=10+E, depending whether there's a carry into the 6th column from the 5th column. Substitute E+1 for A in both those equations and then E can be eliminated so that F=9 or F=8.

Now study the 2nd and 3rd columns. C+D can't be 19, because there's no carry from the rightmost column, and C+D can't be 18 either, because C and D would both have to be 9, yet they must be different. So C+D is 9 or 8, giving no carry. From the third column we see C+1=D, so C+D is odd, and hence equal to 9.

Now we have H=1, F=9, C=4, D=5 and we need A, B, E, and G. From the 4th column from the right we know A+G=11 (not A+G=1 because A is neither 0 nor 1), and of the possible pairs of digits that sum to 11 (2+9, 3+8, 4+7, 5+6) only 3+8 doesn't clash with some digit already assigned. If you let A=8 and G=3 and work out the rest of the letters you'll end with B=5, but 5 is already assigned to D. Switching A and G, A=3, G=8, leads to the solution.