I applied one of the digits 1 through 9 to each cell of the provided net of a cube.
My object was to create a unique 4 digit square number on each face. At the same time I required each vertex to be a 3 digit square. I failed in that objective!
I have 6 unique 4 digit squares but I have duplicated just one of my vertices.

To emulate my "feat":
- a [Magenta] Magenta cell is both the first digit of a 3 and 4 digit square
- an [Orange] Orange cell signifies the first digit of only a 4 digit square, while
- a [Cyan] Cyan cell signifies the first cell only of a 3 digit square.

The digits must be applied to each face by rotation, the direction is defined by need. "A" through "F" represent the 6 faces of the cube while "a" through "h" represent the vertices of the cube when fully assembled. Note: Within the range allowed several squares utilise the same digits, and this is allowed by virtue of the commencement cell.

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But then, there is still the challenge for 6 unique faces and 8 unique vertices.

(In reply to re(3): Clarifications?? by brianjn)

I had been logged on quite a while and had not seen your reply before submitting my last.  Thanks for the clarifications.  Perhaps I am missing something: the six unique faces seem clear, but there are eight vertices, not twelve, where the triads come into action.  I think I can see some simplifications, but there still seem to be a lot of combinations to test. 

Posted by ed bottemiller on 2008-04-30 21:52:09