Replace the same characters by the same numerals so that the mathematical operations are correct.

RE + MI = FA

DO + SI = MI

LA + SI = SOL

LA + SI = SOL implies S = 1. LA + 1I = 1OL would cause L to be 8 or 9, but in both cases, O would have to be 0, which also is in line with DO + SI = MI

A and I can't both be high enough to get L = 8 and carry over into O, so L must be 9, thus 9A + 1I = 109. A + I must thus be 9

From DO + 1I = MI then follows M = D + 1.

So far we have:

RE + MI = FA

D0 + 1I = MI

9A + 1I = 109

A can not be 0, 1, or 9, which are already in use.

A can not be 8, because I would be 1 (used).

A can not be 7, because then I would be 2, and E 5, but that would leave 3, 4, 6 and 8 which could in no way fit D, F, M and R.

A can not be 6, because both I and E would then be 3.

A can not be 5, because that would make I = 4, and thus E is 1 (used).

A can not be 4, because I would become 5, and thus E would be 9 (used).

Assume A = 3, this would imply I = 6, E = 7. Then 2, 4, 5 and 8 would remain. R = 2, D = 4, M = 5, F = 8 will fit.

Assume A = 2, this would imply I = 7, E = 5. Then 3, 5, 6, and 8 would remain. These can not be put for D, F, M and R in any mathematically correct fashion.

So the correct solution would be S = 1, R = 2, A = 3, D = 4, M = 5, I = 6, E = 7, F = 8, L = 9, O = 0

27 + 56 = 83

40 + 16 = 56

93 + 16 = 109

Thanks again,

BoilingOil