$3 \times 3 \times 3$ Magic Cube

The vector space of real $3\times3\times3$ magic cubes is $5$ -dimensional, with basis $\begin{aligned} \mathbf{a} & = \; \left(\begin{array}{ccccccccc} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \end{array}\right) \\ \mathbf{b} & = \; \left(\begin{array}{ccccccccc} 1 & -1 & 0 & -1 & 1 & 0 & 0 & 0 & 0\\ -1 & 1 & 0 & 1 & 0 & -1 & 0 & -1 & 1 \\ 0 & 0 &0 & 0 & -1 & 1 & 0 & 1 & -1 \end{array}\right) \\ \mathbf{c} & = \; \left(\begin{array}{ccccccccc} 0 & -1 & 1 & 0 & 1 & -1 & 0 & 0 & 0 \\ 0 & 1 & -1 & -1 & 0 & 1 &1 & 0 & 0 \\ 0 & 0 & 0 & 1 & -1 & 0 & -1 & 1 & 0 \end{array}\right) \\ \mathbf{d} & = \; \left(\begin{array}{ccccccccc} 0 & 0 & 0 & 1 &-1 & 0 & -1 & 1 & 0 \\ 0 & 1 & -1&-1&0&1&1&-1&0 \\0&-1&1&0&1&-1&0&0&0\end{array}\right) \\ \mathbf{e} & = \; \left(\begin{array}{ccccccccc} 0 &0&0&0&-1&1&0&1&-1\\ -1&1&0&1&0&-1&0&-1&1\\1&-1&0&-1&1&0&0&0&0\end{array}\right) \end{aligned}$ where each successive $3\times3$ block represents the numbers in the bottom, middle and top layer of the cube, respectively. Note that the magic number of the cube $p\mathbf{a} + q\mathbf{b} + r\mathbf{c} + s\mathbf{d} + t\mathbf{e}$ is $3p$ , and its central number is $p$ . We want to know how many ways there are of choosing $p,q,r,s,t$ such that the resulting magic cube only contains the integers from $1$ to $27$ . The magic number of such cubes is $\tfrac19 \times \tfrac12 \times 27 \times 28 = 42$ , so it is clear that $p$ must be equal to $14$ . Then $q,r,s,t$ must be distinct integers chosen from $\pm1,\,\pm2,\,\ldots,\,\pm13$ . A quick computer search of the $4! \times \binom{26}{4} = 358800$ possible choices for $q,r,s,t$ yields a total of $192$ possible magic cubes. Since the rotation group of the cube has order $24$ , this means that there are $192 \div 24 = \boxed{8}$ different magic cubes, to within rotation.

This result is more normally expressed by identifying cubes that can be obtained from each other by reflection, too. This would mean that there were fundamentally only $4$ different magic cubes.

The eight different cubes are given by the choices $(q,r,s,t) = (-13,4,-12,10)$ , $(-12,2,-11,8)$ , $(-10,-2,-7,6)$ , $(-8,-4,-6,5)$ , $(-13,4,-10,12)$ , $(-12,2,-8,11)$ , $(-10,-2,-6,7)$ and $(-8,-4,-5,6)$ , where the last four are the mirror images of the first four.

This result was originally proved by Hendricks in the Journal of Recreational Mathematics in 1972. I have not been able to locate a copy of the paper yet.

There are $27$ unknowns, while there are $27+4$ equations that all equal to $42$ . When that set of linear equations are solved, $22$ of the unknowns are determined by $4$ variables, while the center of the cube is $14$ , which makes a computer search feasible. If we add the condition that all the unknowns are from $1$ to $27$ , then the only values that the center of the faces can have are

$(1, 3, 7, 9, 19, 21, 25, 27)$

which means only the following pairs are possible, if the center $14$ is included:

$(1,27), (3,25), (7,21), (9,19)$

Since any magic cube uses $3$ of the $4$ , there are only $4$ distinct magic cubes possible. However, for this problem, chirals are considered distinct, so double that to get $8$ .

Note: As a check, using all possible distinct values for the 4 variables with the condition that all values are from $1$ to $27$ , there are $192$ possible cases. A cube can have $24$ ways of being rotated, and $2$ chiral versions, so, given that there are only $4$ distinct sets of $3$ pairs described above for the magic cubes, we have $24 \cdot 2 \cdot 4 = 192$ .

FIRST CONSIDER THE THREE PLANES red,green,blue .YOU CAN ASSEMBLE THEM (3-1)! WAYS.THEN YOU CAN UNDERSTAND THAT YOU CAN TAKE THREE PLANES LIKE THAT IN THREE WAYS.SO THE SOLUTION IS 2×2×2=8

The answer is 8 which is the number of magic series for n=3 that sum to ((n^3+1)*(n/2))=42 along all allowed directions. The series A052456 is given at the following links for general value of n:

https://en.m.wikipedia.org/wiki/Magic_series

https://oeis.org/A052456