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Squares on Cubes (Posted on 2008-04-30) Difficulty: 3 of 5
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.
But then, there is still the challenge for 6 unique faces and 8 unique vertices.

See The Solution Submitted by brianjn    
Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
re(2): Clarifications?? | Comment 3 of 21 |
(In reply to re: Clarifications?? by Charlie)

Thanks, Charlie!  This seems a reasonable interpretation, e.g. that "rotation" could be either clockwise or counterclockwise from the midpoint of a side of the cube, and that "defined by need" really means "try both directions -- but ostensibly only one will lead to an overall solution".  I find 19 3-digit squares, and 45 4-digit squares (excluding any with a digit zero). These will probably have to be at least doubled to account for rotating in either direction. The starting point, I assume, will be to decide how many of the "commencement cells" can be identified from the three color clues.  Though he doesn't say so, I assume that a final solution may assume these three color premisses (even though he says he was unsuccessful when he was using them: I guess he means that a "close" solution was available, but that two of the vertex sets were duplicates.) It is not clear whether a 4-digit square and its digital reverse would each count as a "unique" solution; I am not sure whether his penultimate Note is intended to address that matter (or any other -- it's too vague). I hope he has ascertained and verified that there is indeed a solution meeting (some) "uniqueness" criterion.
  Posted by ed bottemiller on 2008-04-30 16:48:48

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