Previously we saw a strategy to solve this puzzle (each letter stands for one and only one number)
FIVE - FOUR = ------ ONE + ONE = ------ TWO
First we found a comfortable representation of the space of possible solutions, a representation we could handle with paper and pencil: to do this paradoxically we had to think in an algorithmic manner, as if we were going to instruct a computer to solve the problem – a very limited computer indeed!…
Let’s get started
One can rewrite the operations composing the statement of the puzzle to make the task a little bit easier. For instance rewriting the subtraction
FOUR + ONE = ------ FIVE
Therefore the last digit of R + E must equal E. For this to be true R = 0. From the addition
ONE + ONE = ----- TWO
Now the rightmost column says that O is the rightmost digit of E + E. Hence O must be one of 2, 4, 6, 8 (not 0, recall our earlier assumption), since TWO is an even number. But from the leftmost column of this sum we know that either O + O = T or (in case of carry from the nearby column) O + O + 1 = T, so that O cannot be greater than 4 (otherwise there would be a carry from the leftmost column, so that the result would be four digits long instead of three). Therefore either O = 2 or O = 4. We can use the rightmost column of the sum to infer that in the former case either E = 1 or E = 6, in the latter either E = 2 or E = 7.
Indeed if O=2
1 + 6 + 1 = and 6 = ---- ---- 2 12 (we keep only the lsd 2)
Likewise for O=4
2 + 7 + 2 = e 7 = ---- ---- 4 14 (we keep only the lsd 4)
In turn we see from the addition
ONE + ONE = ----- TWO
that when O = 2 we can have either T = 4 or T = 5 (according to whether there is carry or not); while O = 4 implies that either T = 8 or T = 9 (again, carry).
To summarize what we’ve found
Now let’s go back to the subtraction: since R = 0 we can write it as follows (just “forgetting” the rightmost column)
FOU + ON = ----- FIV
The middle column says that either O + O = I or O + O + 1 = I (in case of carry, i.e. U + N >9). But wait a moment! We also had that either O + O = T or O + O + 1 = T. Since each letter represents a unique digit T and I cannot e equal. Let’s write this with a table (using what we know about O)
Put this into the larger table to obtain
This thing is becoming a bit large, but it enables us to keep the calculation down to a manageable size. I also added a column with the spare digits (i.e. digits we haven’t associated to a letter yet). Recall that we have the letters W, N, U, V, F yet to determine.
Now it’s time enough for another pause to reflect a little bit. I invite you to go on the path shown in this post, at least for a few steps. The end of the story in a following post.