# “8 Dice arranged as a Cube” Face-Sum Problem

I found this here:

Sum Problem

Given eight dice. Build a $2\times 2\times2$ cube, so that the sum of the points on each side is the same.

$\hskip2.7in$

Here is one of 20 736 solutions with the sum 14.
You find more at the German magazine "Bild der Wissenschaft 3-1980".

Now I have three (Question 1 moved here) questions:

1. Is $14$ the only possible face sum? At least, in the example given, it seems to related to the fact, that on every face two dice-pairs show up, having $n$ and $7-n$ pips. Is this necessary? Sufficient it is...

2. How do they get $20736$? This is the dimension of the related group and factors to $2^8\times 3^4$, the number of group elements, right?

i. I can get $2^3$, by the following: In the example given, you can split along the $xy$ ($yz,zx$) plane and then interchange the $2$ blocks of $4$ dice. Wlog, mirroring at $xy$ commutes with $yz$ (both just invert the $z$ resp. $x$ coordinate, right), so we get $2^3$ group lements.  ii. The factor $3$ looks related to rotations throught the diagonals. But without my role playing set at hand, I can't work that out.  iii. Would rolling the overall die around an axis also count, since back and front always shows a "rotated" pattern? This would give six $90^\circ$-rotations and three $180^\circ$-rotations, $9=3^2$ in total. $$\\$$ Where do the missing $2^5\times 3^2$ come from?

3. Is the reference given, online available?

EDIT

And to not make tehshrike sad again, here's the special question for $D4$:

What face sum is possible, so that the sum of the points on each side is the same, when you pile up 4 D4's to a pyramid (plus the octahedron mentioned by Henning) and how many representations, would such a pyramid have?

Thanks

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related: math.stackexchange.com/q/154451/19341, splitted due to request here: math.stackexchange.com/a/229592/19341 –  draks ... Nov 6 '12 at 7:27
The point is, you really should ask one question per question. –  Phira Nov 6 '12 at 8:25
For the bonus question "(ok, 5 in the center for stability)", note that if you pile up four tetrahedra to form a larger tetrahedron, the open space in the middle is not in the shape of a tetrahedron, but is an octahedron. So putting a fifth tetrahedron in the middle is not going to stabilize the construction appreciably. –  Henning Makholm Nov 6 '12 at 11:59
@HenningMakholm thanks +1 for the correct stabilizer. –  draks ... Nov 6 '12 at 13:02