0
$\begingroup$

A magic square is one in which the sum of the numbers in each row, column, and both main diagonals is the same. The numbers in the magic square can only be 1 to 9.

a 3x3 magic square example: a 3x3 magic square example

There are 8 combinations of a 3x3 magic square.

A white Rubik's cube has 6 sides. There are 9 spaces on each side for numbers from 1 to 9, without color. a white Rubik's cube

If we fill each side of a white Rubik's cube with a magic square, how many combinations can be filled on this Rubik's Cube?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

I assume that each magic square must be composed of numbers $1,2,\dots,9$ (or $1,2,\dots, n^2$ in general), and that under filling a cube with magic squares we understand assigning a number to each $1\times 1\times 1$ cell at the cube surface such that the numbers at each face form a magic square.

Filling $3\times 3\times 3$ cube with magic squares is not possible. Notice that the center number of a magic square is $5$, and once a number at a corner of a magic square is fixed to be $a\ne 5$, the number at the opposite corner (ie. on the same diagonal as $a$) is $10-a\ne a$. However, in the cube with corners having coordinates $\pm 1$ (say), there is a triangle with vertices $(-1,-1,-1)$, $(-1,1,1)$, $(1,-1,1)$, whose sides are formed by surface (magic) diagonals, which is impossible.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thanks for your answer. I have clarified my question. $\endgroup$
    – Chuck Liu
    Commented Jan 14, 2023 at 23:39
  • $\begingroup$ @ChuckLiu: I don't see much clarification with respect to cube filling. Does my interpretation match your intended question for $n=3$? $\endgroup$
    – Max Alekseyev
    Commented Jan 15, 2023 at 16:48
  • $\begingroup$ I assume the six sides (up, down), (left, right), (forward, backward). For (up, down), we can get 8 combinations. So I think the number of combination is between from $8$ to $8^3$. $\endgroup$
    – Chuck Liu
    Commented Jan 18, 2023 at 5:58
  • $\begingroup$ @ChuckLiu: Then magic squares become irrelevant. Think of them as colors and thus one need to enumerate independent colorings of the cube faces. Solution to this problem is well known, even if we count colorings up to rotations of the cube. $\endgroup$
    – Max Alekseyev
    Commented Jan 18, 2023 at 14:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .