## Different Totals

#### Problem

Using each of the digits 1 to 5 once, it is possible to place them in the grid so that the row and column have the same total.

How many different totals can this be done with?

#### Solution

Begin by listing all possible totals using three digits from 1, 2, 3, 4, 5:

Total | |||

1 | 2 | 3 | 6 |

1 | 2 | 4 | 7 |

1 | 2 | 5 | 8 |

1 | 3 | 4 | 8 |

1 | 3 | 5 | 9 |

1 | 4 | 5 | 10 |

2 | 3 | 4 | 9 |

2 | 3 | 5 | 10 |

2 | 4 | 5 | 11 |

3 | 4 | 5 | 12 |

So it is possible to fill the grid with three different totals:

8 = 2 + 5 + 1 = 1 + 3 + 4 9 = 5 + 1 + 3 = 3 + 2 + 4 10 = 1 + 4 + 5 = 5 + 2 + 3 | i.e. |

Alternatively, note that the common square in the top right corner is always odd... The sum of all five digits, 1 + 2 + 3 + 4 + 5 = 15. If an even digit is placed in the common square: 15 2 = 13 or 15 4 = 11, the remaining total will be odd, and this cannot be split between the two legs equally. Hence the only way in which this can be completed is to place an odd digit in the common square: 15 1 = 14 (7 on each leg), 15 3 = 12 (6 on each leg) and 15 5 = 10 (5 on each leg).

What about using the digits 1 to 7 with the following grid?