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Hypercells (Posted on 2009-02-02) Difficulty: 4 of 5
The network represents a map of the edges and vertices of a hypercube or more precisely, a tesseract when imposed within 2D space.

The orange squares represent the faces of the internal cell, a cube in 3D space, and is defined by vertices A to H. The larger bolder letters, I to P are the vertices of the outer cell/cube.

The diagonal lines are the edges which link the inner to the outer forming the faces that configure the cells which interface to the inner and outer cells.

Let each hypercell have a value which is the vertex sum of each of its enclosing faces (or three times the sum of its 8 vertices).
The vertices are to be numbered uniquely from 1 to 16, and for consistency of reference, let A always be 1.
Vertex values:
ABCD
EFGH
IJKL
MNOP
= 136
Cell 1: ABCDEFGH
Cell 2: IJKLMNOP
Cell 3: ADFGILMP
Cell 4: ABEFIJMN
Cell 5: ABCDIJKL
Cell 6: BCEHJKNO
Cell 7: CDGHKLOP
Cell 8: EFGHMNOP
Find sets of values for A through P so that all the cells have the same value.

Note: while the on-line calculator may be useful, a spreadsheet should prove more valuable for those able to use one.

See The Solution Submitted by brianjn    
Rating: 4.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
34 | Comment 6 of 10 |
Daniel attests to cell values of 204.

The sum from 1 to 16 is 136. 

              Both are multiples of 34!

Given that thought, I mean 34, does the 4 x 4 "magic square" have any relevance to this?

Er?  Now that I raise this, how do "magic square" solutions fit his problem?

Have I by accident given that concept a "new life"?  
Wonder? 3 x 3 = 15!  Good thought.  Should be but one answer once transformations are considered.





  Posted by brianjn on 2009-02-03 09:10:26
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