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Challenge 8 Answer


On the tic-tac-toe board

Solution submitted by the user Valerio Deo.

There are currently many references to the solution of so-called magic squares (this is the "name of the game"), but here, we will use intuitive reasoning to solve this challenge:

Our first concern will be to find groups of 3 distinct digits whose sum is 15. The process should be as natural as possible, consisting of organizing families from the smallest to the largest.

  • Starting with the family of 1, we could think of 2 for the next element of the group, but 12 would still be left to reach the sum 15. So the second digit must be 5 so that the third one is as large as possible, or 9 so to have the sum 15. With this procedure we get the family of digit groups starting with 1:

    159
    168
    The family of 1 only has 2 groups and it was not possible to use the digits 2, 3, 4, 7.

  • The next family will be from groups starting with 2. and the other two members should add 13:

    249
    258
    267
    Unused numerals: 1, 3.

  • Family of 3:

    348
    357
    Unused figures: 1, 2, 6, 9.

  • Family of 4:

    429
    438
    456
    Unused numerals: 1, 7.

  • Family of 5:

    519
    528
    537
    546
    The 9 digits were used!

  • Family of 6:

    618
    627
    645
    Unused numerals: 3, 9.

  • Family of 7:

    726
    735
    Unused figures: 1, 4. 8, 9.

  • Family of 8:

    816
    825
    834
    Unused numerals: 7, 9.

  • Family of 9:

    915
    924
    Unused numerals: 3, 6. 7, 8

The configuration of the so-called "Tic Tac Toe" is known as the 3 × 3 Matrix, that is, an interlaced set of 3 rows and 3 columns forming a "square" with 9 cells. In the present case, the 9 digits must occupy the 9 cells in such a way that, in any row, any column or any diagonal, the sum of the 3 digits is always 15, forming the so-called 3 × 3 Magic Square

Considerations for the 3 × 3 Magic Square whose sum is 15:

  • The central cell belongs simultaneously to the central line, the central column and the two diagonals, forming four groups of digits where one of them is common to all.

  • The family of 5 is the only one that brings together 4 groups of numbers, which leads us to conclude that the number 5 should occupy the central position of the matrix:

    5

  • By observation, we found that there are 4 families with 3 groups (2, 4, 6, 8) and 4 families with 2 groups (1, 3, 7, 9). In any case, there is always a group containing the digit 5.

  • We also observe that from the vertex cells of the square, 3 groups of digits are always generated, occupying a row, a diagonal and a column. Thus, the families of 3 groups, ie 2, 4, 6 and 8, should occupy such positions:

    24
    5
    68

  • It remains for us to "fit" the families of 2 groups, that is, 1, 3, 7 and 9 into the still empty cells, taking care to verify, in each case, if the sum with the other digits of the same row or column. totals 15:

    294
    753
    618

The above result would be a fully satisfactory response to the proposed challenge. But we must still consider some other possibilities.

The fact that we chose the first vertex for the position of the number 2 was of pure convenience because we could choose any of the other vertices to start the reasoning.
Geometrically, choosing the other vertices means promoting a "rotation" in the matrix where the axis of rotation would be perpendicular to the paper. Let us then choose the counterclockwise direction for successive 90 degree rotations. This way we get 3 more possible solutions:

438
951
276
816
357
492
672
159
834

Take the first solution and imagine another kind of rotation in which the axis would now be vertical, pertaining to the paper plane and, say, passing through the center of the matrix. Let's promote a 180 degree rotation (the numbers remain as they are):

492
357
816

If in this new solution we promote 3 more 90 degree rotations with the axis perpendicular to the paper plane, we will find 3 more possible solutions:

276
951
438
618
753
294
834
159
672

Answer:

By bringing together all the above solutions we will have a set of 8 magic squares as a solution to the proposed challenge:

294
753
618
438
951
276
816
357
492
672
159
834
492
357
816
276
951
438
618
753
294
834
159
672

Final grade:

We could still think about promoting a horizontal axis rotation, but in 2D, as we would see, the solutions would be redundant, that is, they would coincide with the solutions already found.

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