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 comfor*table* 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

R | O | T | E |
---|---|---|---|

0 | 2 | 4 | 1 |

0 | 2 | 4 | 6 |

0 | 2 | 5 | 1 |

0 | 2 | 5 | 6 |

0 | 4 | 8 | 2 |

0 | 4 | 8 | 7 |

0 | 4 | 9 | 2 |

0 | 4 | 9 | 7 |

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*)

O | T | I |
---|---|---|

2 | 4 | 5 |

2 | 5 | 4 |

4 | 8 | 9 |

4 | 9 | 8 |

Put this into the larger table to obtain

R | O | T | I | E | spare |
---|---|---|---|---|---|

0 | 2 | 4 | 5 | 1 | 36789 |

0 | 2 | 4 | 5 | 6 | 13789 |

0 | 2 | 5 | 4 | 1 | 36789 |

0 | 2 | 5 | 4 | 6 | 13789 |

0 | 4 | 8 | 9 | 2 | 13567 |

0 | 4 | 8 | 9 | 7 | 12356 |

0 | 4 | 9 | 8 | 2 | 13567 |

0 | 4 | 9 | 8 | 7 | 12356 |

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.