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Fill the following 3X5 table with numbers from 1 to 15 such that each column has the same sum and each row has the same sum. Each number should be used only once.


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Edited by brhan
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Posted · Report post

Fill the following 3X5 table with numbers from 1 to 15 such that each column has the same sum and each row has the same sum. Each number should be used only once.


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| 10| 7 | 2 | 6 | 15|
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| 11| 5 | 9 | 14| 1 |
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| 3 | 12| 13| 4 | 8 |
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-------------------------------| 3 | 5 | 13 | 4 | 15 |-------------------------------| 10 | 7 | 9 | 6 | 8 |-------------------------------| 11 | 12 | 2 | 14 | 1 |-------------------------------

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This is the way I see it.

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| 3 | 5 | 13 | 4 | 15 |

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| 10 | 7 | 9 | 6 | 8 |

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| 11 | 12 | 2 | 14 | 1 |

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Posted · Report post

Don't know if there is more than one answer (other than rearranging columns or each row), but here it is:

..1.11.14.12..2

.15..3..4..5.13

..8.10..6..7..9

We know what the sum of each column must be = 24. Why? It is the average sum of any three numbers (first + last + mid) = (1 + 15 + 8), which also happens to be my first column.

We know the sum of each row must be = 40. Why? Again, the average sum of any 5 numbers. (first + last + mid + mid-of-first-half + mid-of-last-half) = (1 + 15 + 8 + 4 + 12) ... Note that this cannot be a row, since 15 and 8 are already in a column.

Finally, we can set at least one row. I chose to fix the "highest" row, which I assumed to be the one starting at 15. The highest numbers (since we are at an average sum) should be paired with the lowest. However, 15 cannot be paired with 2, 3, 4, since it doesn't allow enough to add to 40. Thus, it becomes 15,3,4,5, with 13 as the last number. Once those are set, the rest is really just easy.

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