Given A = 5 and O = 4, we have
I + A + O = I + 9
and the only way for I + 9 ≡ I (mod 10) is if a 1 is carried over from the sum in the units columns, since that would make
I + 9 + 1 ≡ I + 10 ≡ I (mod 10)
which means
10 ≤ 2C + E ≤ 19
Now, I ∈ {0, 1, 2, 3, 6, 7, 8, 9}, so we'll have to carry another 1 into the next column's sum.
I + A + O + 1 = I + 10 ===> 3T + 1 = W
This tells us that
3T = W - 1
i.e. W - 1 must be a multiple of 3. Then either W = 4 or W = 7, but the first case is dropped because we already have O = 4. So we know W = 7 and T = 2.
From here, I think it's a matter of checking all the possible cases. Work your way up from the smallest value for I, and determine what possible values N can take.
• Suppose I = 0. Then N ∈ {1, 3, 6, 8, 9}.
•• If N = 1, then 2C + E = 11. Consider the possible integer partitions of 11 that involve an even number (the contribution of 2C):
10 + 1 ===> C = 5, E = 1
8 + 3 ===> C = 4, E = 3
6 + 5 ===> C = 3, E = 5
4 + 7 ===> C = 2, E = 7
2 + 9 ===> C = 1, E = 9
None of these options work, since each of {1, 2, 4, 5, 7} are taken.
•• If N = 3, then 2C + E = 13, which means
12 + 1 ===> C = 6, E = 1 (bingo!)
10 + 3 ===> C = 5, E = 3
8 + 5 ===> C = 4, E = 5
6 + 7 ===> C = 3, E = 7
4 + 9 ===> C = 2, E = 9
The other options don't work, since each of {2, 3, 4, 5, 7} are taken.
So, the least value is WIN = 703 :
TIC = 206
TAC = 256
TOE = 241
TIC + TAC + TOE = 703
