While waiting at the Spaceport on Zargon for my long flight back to Earth from vacation, I did some souvenir shopping and found some strange looking dice used for weird board games known only to the Zargonians. These dice, each identical, had four sides instead of six, with each face arrayed into four triangular sections. While studying the dice, I noticed a few interesting things that might relate to how they're used in a game.

Each triangular section was numbered variously 1 through 9, with every digit except 1 and 9 appearing twice on each die (but on different faces). On each face, the center space was the sum of two of its corners, except on one face where it was the difference, and the centers comprised 4 consecutive numbers. The corners on one face summed to a square, on another to a prime number and the corners on the remaining two faces were each comprised of consecutive numbers.

I also noticed that only three odd numbers occurred around two of the vertices while only even numbers occurred around the other two, and in each case the three numbers were different. The product of one set of even numbers was a cube, the sum of the other three even numbers also equalled the sum of the digits in their product, and only one odd number was duplicated between the two odd vertices. Further, the sums of the three numbers at each vertex result in 4 more consecutive numbers.

What were the numbers on each face and how were they arranged?
--------------------------------------------------------------
For convenience, present the solution as an unfolded die thusly. Your arrangement may vary from the solution (what you indicate as the center face, for example), but the contents and relationships of the faces won't change.

             _
            /*\
           / * \
          /*   *\
         /-------\
        /*\*   */*\ 
       / * \ * / * \	    
      /*   *\*/*   *\

here is my solution and I believe I have determined that it is unique barring obvious rotations

          /9\
        / 8  \
       /2    6\
      ---------
     /8\4   2/4\
    / 6 \  5 / 7 \
   /1   7\3/5   3\
   ---------------

in case it is difficult to see from my solution the center faces

are 5,6,7,8

 

now to describe my reasoning process

I started by using Excel to list all possible combinations of 3 numbers from 1 to 9 without repetition.  I then eliminated all combinations that were not consecutive or did not sum up to either a square or a prime.  After that I sorted them into 3 groups; consecutive, square, and prime depending on which of the 3 they satisfied, there was one combination that appeared twice 2,34, wich is both consecutive and sums to a square.  I then printed this out for future use.

next I found all combinations of 3 even numbers of which there are only a few.  And only 1 had its product being a cube so the only possibility for that vertex is 2,4,8.  Also only 1 other combination of even numbers had its sum equal to the sum of the digits of its product and that was 2,4,6.  Now the sum of each of these is 14 and 12 respectively.  Since the sum of the vertices is consecutive that means that 13 must be the sum of one of the odd vertex and since the other one must sum to an odd number that sum must be 15 to complete the series.

now there are 2 combinations of 3 odds each that sum to 13 and 15 but looking at these only 1 combination of the 4 share only 1 common number and that is 1,3,9 for 13 and 3,5,7 for 15

so now we know the combination of the vertices

2,4,8

2,4,6

1,3,9

3,5,7

now I arbitrarily placed the 2,4,8 combination as in my solution and proceeded to make several deductions from there

now for brevity I will no place my complete list of deductions but will summarize them.  I went through each possible combination of placement of the vertices and their rotations and only one of them resulted in a face that sumed to a square, a prime and the other 2 consecutive and that is the solution above.

Posted by Daniel on 2008-11-30 01:37:44