An alphametic is a type of number puzzle. It involves a sum (or difference, or product, or some other arithmetical operation) in which each digit has been replaced by a distinct letter of the alphabet. Perhaps the most famous one was one of the original ones, published by Henry Dudeney in 1924:

S | E | N | D | |

+ | M | O | R | E |

M | O | N | E | Y |

This page will discuss how to solve alphametics, and then present several alphametic puzzles.

To illustrate how to solve alphametics, we'll take the example above:

S | E | N | D | |

+ | M | O | R | E |

M | O | N | E | Y |

In an alphametic, no number typically is allowed to begin with 0, so neither S nor M can be 0. Now, if you add two four-digit numbers, the result must be no more than 19,998, so the only possibility for M is **M = 1**.

Now, let's look at the thousands place. We now have S + 1 = O. We know that there is a carry into the ten thousands place (so really S + 1 = 10 + O), and there may or may not be a carry into the thousands place. So, there are three possibilities for this column:

- There is a carry into the thousands column and S = 8, making O = 0.
- There is no carry into the thousands column and S = 9, making O = 0.
- There is a carry into the thousands column and S = 9, making O = 1.

However, O cannot be equal to 1, since M = 1 and each letter is a different digit. Therefore, **O = 0**. Here's what we have now:

S | E | N | D | |

+ | 1 | 0 | R | E |

1 | 0 | N | E | Y |

Now, let's look at the hundreds column, which is E + 0 = N. Now, since each letter must represent a different number, there must be a carry into the hundreds column, and N must be equal to E + 1. Now, E can't equal 9 and N equal 0, because another letter, O, is 0). This means that there is no carry into the thousands column, so **S = 9**.

Now, let's look at the tens column. We have N + R = E. We also know there is a carry out of the tens column, so there are two possibilities here:

- There is a carry into the tens column, so 1 + N + R = 10 + E
- There is no carry into the tens column, so N + R = 10 + E

Since we know that N = E + 1, these two possibilities are equivalent to:

- There is a carry into the tens column, so 1 + N + R = 9 + N or R = 8
- There is no carry into the tens column, so N + R = 9 + N or R = 9

However, R cannot equal 9 because S = 9, so **R = 8**.
We now have:

9 | E | N | D | |

+ | 1 | 0 | 8 | E |

1 | 0 | N | E | Y |

Now, let's look at the ones column. Since R = 8, there must be a carry out of the ones column into the tens column. This means that D + E ≥ 10. The ones digit of this sum (which equals Y) can't be 0 or 1 though (they're both taken) so D + E ≥ 12. Since neither can be 8 or 9, one of them must be 7 and the other must be 5 or 6. Now, N = E + 1, so N must be either 6 or 7. The only possibility that works is that **D = 7, E = 5, N = 6**. Since D + E = 12, that means that **Y = 2** and the problem is solved:

9 | 5 | 6 | 7 | |

+ | 1 | 0 | 8 | 5 |

1 | 0 | 6 | 5 | 2 |

So, the resulting sum is 9567 + 1085 = 10652, which is correct.

Here are some other alphametics you can try. If you'd like to try the first two interactively, you can find them on All Fun and Games (alphametic #1, alphametic #2).

1.

F | O | R | T | Y |

T | E | N | ||

+ | T | E | N | |

S | I | X | T | Y |

2.

D | O | N | A | L | D | |

+ | G | E | R | A | L | D |

R | O | B | E | R | T |

3.

F | E | A | R | |

+ | R | A | G | E |

G | R | I | E | F |

4.

A | B | C | D | E |

× | 4 | |||

E | D | C | B | A |

5. (Note: There are multiple possible solutions)

T | W | O | |||

× | S | I | X | ||

T | W | E | L | V | E |

6.

A | N | T | E | |

− | E | T | N | A |

N | E | A | T |

The answers are on the answers page.