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Consider the following image.
Within the grid the are a total of 19 cells.
We have one cell for zero leaving 18 cells.
Shading nine cells we create 2 sets of the digits 1 to 9. With one set being on the shaded cells and the other set being on the unshaded cells.
The shaded digits within all 15 straight lines running through the grid sum to the exact same total as the unshaded digits within the same line.

Magic Hexagon with equal sum lines

Question 1
Is there a solution for all of the 19 different digits being the centre cell?
Example: In the above image the centre cell is a shaded 3 so is it possible to put a shaded 9 in the centre?

Question 2
Ignoring rotations and reflections, how many different ways are there of arranging the numbers within the grid so that the shaded cells are equal to the unshaded cells in all 15 lines running through the grid?

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2 Answers 2

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Edit: while preparing a complete set of possibilities I found more duplicates:

With rotations, reflections and colour reversals I find 19 unique solutions:
14 solutions with 0 in the middle
1 solution with 1 in the middle
1 solution with 2 in the middle
1 solution with 3 in the middle
2 solutions with 6 in the middle

enter image description here

Below is the original set of my 0-centre solutions, rearranged to show duplicates.
These are the two pairs on the first row, which are the same with colour reversal.

enter image description here


Previously: I independently got a similar but slightly different answer from @user1502040

There is one solution each with 1, 2 or 3 in the middle
There are two solutions with 6 in the middle

      4  -9   5
    1   3   0  -4
 -5  -2   6   2  -1
    8  -8  -7   7
     -3   9  -6

      4   0  -4
    3  -9  -3   9
 -7   1   6   5  -5
    8   2  -8  -2
     -1  -6   7
 

I found 16 solutions with 0 in the middle
Here are three of them

       1  -6   5
     4   2  -9   3
  -5   7   0   6  -8
    -3  -4  -2   9
       8  -7  -1

       1  -6   5
    -9   2   4   3
   8   7   0  -7  -8
    -3  -4  -2   9
      -5   6  -1

       1  -7   6
    -9   4   3   2
   8   5   0  -5  -8
    -2  -3  -4   9
      -6   7  -1
 

So that's the same number of solutions.

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    $\begingroup$ Interesting thst only 6 and its factors can take the centre stage $\endgroup$
    – Maff
    Mar 12, 2023 at 7:39
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Question 1:

It is only possible to put the numbers 0, 1, 2, 3, and 6 in the center (or their negations, by symmetry). Here are some corresponding solutions:

     -9   1   8
    3   4  -5  -2
  6  -7   0   7  -6
    2   5  -4  -3
     -8  -1   9

      5  -9   4
   -7   0   8  -1
  2   6   1  -6  -3
    3  -8  -2   7
     -5   9  -4

      1   4  -5
    7  -9   6  -4
 -8   0   2  -3   9
    5  -6   8  -7
      3  -1  -2

      1   7  -8
    6  -9   4  -1
 -7   0   3  -5   9
    2  -4   8  -6
      5  -2  -3

      4   0  -4
    3  -9  -3   9
 -7   1   6   5  -5
    8   2  -8  -2
     -1  -6   7
 

Question 2:

I believe there are 26 solutions up to symmetry. I can list them if anyone is interested.

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  • $\begingroup$ If you could list them all that would be great. $\endgroup$
    – Maff
    Mar 12, 2023 at 7:35
  • $\begingroup$ Are you sure negative numbers are allowed? $\endgroup$ Mar 13, 2023 at 8:21
  • $\begingroup$ @PlaceReporter99 Negative means shaded. $\endgroup$ Mar 13, 2023 at 9:49

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