The most important thing is the relation between the number of small boxes and the sum of these boxes.

If the sum of 2 boxes is 3, then the only combination is 1+2. You have to remember the all deterministic combinations.

The following table contaions all deterministic combinations.

number of boxes | sum | combination |
---|---|---|

2 | 3 | 1+2 |

4 | 1+3 | |

16 | 7+9 | |

17 | 8+9 | |

3 | 6 | 1+2+3 |

7 | 1+2+4 | |

23 | 6+8+9 | |

24 | 7+8+9 | |

4 | 10 | 1+2+3+4 |

11 | 1+2+3+5 | |

29 | 5+7+8+9 | |

30 | 6+7+8+9 | |

5 | 15 | 1+2+3+4+5 |

16 | 1+2+3+4+6 | |

34 | 4+6+7+8+9 | |

35 | 5+6+7+8+9 | |

6 | 21 | 1+2+3+4+5+6 |

22 | 1+2+3+4+5+7 | |

38 | 3+5+6+7+8+9 | |

39 | 4+5+6+7+8+9 | |

7 | 28 | 1+2+3+4+5+6+7 |

29 | 1+2+3+4+5+6+8 | |

41 | 2+4+5+6+7+8+9 | |

42 | 3+4+5+6+7+8+9 | |

8 | 36 | 1+2+3+4+5+6+7+8 |

37 | 1+2+3+4+5+6+7+9 | |

38 | 1+2+3+4+5+6+8+9 | |

39 | 1+2+3+4+5+7+8+9 | |

40 | 1+2+3+4+6+7+8+9 | |

41 | 1+2+3+5+6+7+8+9 | |

42 | 1+2+4+5+6+7+8+9 | |

43 | 1+3+4+5+6+7+8+9 | |

44 | 2+3+4+5+6+7+8+9 | |

9 | 45 | 1+2+3+4+5+6+7+8+9 |

I teach you how to remember this table.

When number of boxes is less or equal to 7, each number of boxes have 4 deterministic combinations. When number of boxes is 8, there are 9 combinations. When number of boxes is 9, there is 1 combination (sum from 1 to 9).

In the case of 9 boxes, you have to use all digits (9 kinds). This is the only solution.

In the case of 8 boxes, you don't use one digits. If you don't use 1, then 45-1 = 44 is the sum.(2+3+4+5+6+7+8+9=44). If you don't use 2, then 45-2 = 43 is the sum.(1+3+4+5+6+7+8+9=43). So, 45 - sum is the number that you don't use.

Other cases (from 2 boxes to 7 boxes), there are always 4 combinations in each case. The 2 combinations are big and the rest 2 combinations are smalls.

Let's find the smallest 4 boxes combination. The addition of the sequentical numbers from 1 gives the smallest sum. So, 1+2+3+4=10 is the smallest.

The smallest sum of 4 boxes is 10. Now, let's think about the next number 11. To get the combination, you have to increase one of the numbers 1+2+3+4. If you increase 1 or 2 or 3, then the combinations is 2+2+3+4, 1+3+3+4, 1+2+4+4. But these combination is bad because same number appears more than once. The only combination is increase the biggest nuber 4 by one. So, the possible combination is 1+2+3+5 = 11.

Let's think about the next case : the sum of 4 boxes is 12. But in this case, there are two combinations, 1+2+3+6 and 1+2+4+5. You have to use 1 and 2. The rest two boxes are 3+6 or 4+5.

There are 4 boxes.

- 1 (upper)
- Horizontal: sum of 2 boxes is 3, the only combination is 1+2.
- Vertical: sum of 2 boxes is 4, the only combination is 1+3.
- The common number is 1.
- 1 (lower)
- Horizontal: sum of 2 boxes is 4, the only combination is 1+3.
- Vertical: sum of 3 boxes is 7, the only combination is 1+2+4.
- The common number is 1.
- 9
- Horizontal: sum of 2 boxes is 18, the only combination is 8+9.
- Vertical: sum of 2 boxes is 16, the only combination is 7+9.
- The common number is 9.
- 4
- This is a little difficult.
- Horizontal: sum of 4 boxes is 10, the only combination is 1+2+3+4.
- Vertical: sum of 4 boxes is 34, the only combination is 4+6+7+8+9.
- The common number is 4.

The very useful hint: if one direction is the combination of big numbers and other direction is the combination of small numbers (like the case of 4), the maximum number of small numbers and the minimum number of big numbers is very often the same.

Let's fill some other boxes.

A little difficult case is neibors of 4(cyan). The vertical combination is 1+2+4 and 4 is used. then the rest is 1+2. So, the upside box of 4 is 1 or 2. As the sum of horizontal 2 boxes is 4, 1+3 is the only combination. The common number of 1+3(horizontal) and 1+2+4(vertical) is 1.

You can find 5 (cyan). Horizontal is 5+7+8+9, Vertical is 1+2+3+5. The common number is 5.

I only show you very very fundamentail strategy, but you can fill many boxes.

to be continued...

Index of Strategy