In the following octahedral net the vertices A, B, C, D, E and F, and are to be assigned distinctive values from 1 through 9.


M, N, O, P, Q, R, S and T are the values of the respective sums of the three vertices which form the respective surfaces to which each is assigned.

M + N = O + P = Q + R = S + T

Find unique sets of values¹ for A, B, C, D, E and F such that the values of M through T (but not necessarily in that order) form series which increment by 2 of which there are 7.

Background statistics:
                                   Unique        Actual
                M+N=   Interval     Sets2       Solutions
                 21       1           1            96
                 23       1           2            96
                 25       1           2            96
                 26       2           2            96
                 27       1           5           480
                 29       1           4           288 
                 30       2           3           144
                 31       1           8           288
                 33       1           6           480
                 34       2           2            96
                 35       1           2            96
                 37       1           2            96
                 39       1           1            96
                                     40          2448 

Note:
1. For M+N=21,
          1, 2, 3, 6, 4, 5
          1, 2, 3, 6, 5, 4
      and 1, 2, 5, 3, 4, 6  
   are the first 3 of 96 solutions.  The 96 values do not
   discriminate amongst vertex rotation or reflections. 
2. Each unique set configures the octahedron in more ways than one.

Initially, the problem states there are 7 possible series for the eight values M through T to increment, in some order, by 2.  As I see it, there are only 5 such possible series.  The largest any of these values can be is 24 (consisting of vertices 7-8-9), while the smallest can only be 6 (consisting of 1-2-3).  The five series for M through T would then be (with some possible face combinations indicated in brackets):

 

10, 12, 14, 16, 18, 20, 22, 24 (7-8-9 only) = 136 (being the octrahedral total of M through T, with M+N, O+P, etc. = 34 for each pair)

 

9 (2-3-4, 1-2-6 or 1-3-5), 11, 13, 15, 17, 19, 21, 23 (6-8-9 only) = 128 (= 32 each pair)

 

8 (1-3-4 or 1-2-5), 10, 12, 14, 16, 18, 20, 22 (6-7-9 or 5-8-9) = 120 (= 30 each pair)

 

7 (1-2-4 only), 9, 11, 13, 15, 17, 19, 21 (6-7-8, 5-7-9 or 4-8-9) = 112 (= 28 each pair)

 

6 (1-2-3 only), 8, 10, 12, 14, 16, 18, 20 (5-7-8 or 4-7-9) = 104 (= 26 each pair)

 

The various pairs (M+N, O+P, etc., in no particular order) can also then be determined for each series, to comprise the first + last, 2^nd + 7^th, 3^rd + 6^th, and 4^th + 5^th numbers in the series.

 

Now, each of A through F will factor into 4 different faces (i.e. 4*1= 4, 4*2=8, 4*3=12,…, and/or 4*9=36).  The above five possible series totals of 136, 128, 120, 112 or 104 will therefore represent the sums of six of the above nine '4*' possibilities.  For example, (4*1) + (4*3) + (4*4) + (4*5) + (4*6) + (4*7) would sum to 104, but 1, 3, 4, 5, 6 and 7 can't represent A through F since we know the first 6 in that particular series requires a 2!  (4*1) + (4*2) + (4*3) + (4*7) + (4*8) + (4*9) would sum to 120, but again the final 22 in that series requires a 5 or 6, so 1, 2, 3, 7, 8, 9 are also out for A through F!

 

Beyond that, I think it's spread-sheet time (definitely not my forte), so I'll have to leave that to others!

Edited on December 30, 2008, 5:51 pm

Edited on December 30, 2008, 5:54 pm

Posted by rod hines on 2008-12-30 17:41:36