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Because of the factorization (2n + 1)(n 2 + n + 1), it is impossible for a centered cube number to be a prime number. [3] The only centered cube numbers which are also the square numbers are 1 and 9, [4] [5] which can be shown by solving x 2 = y 3 + 3y, the only integer solutions being (x,y) from {(0,0), (1,2), (3,6), (12,42)}, By substituting a=(x-1)/2 and b=y/2, we obtain x^2=2y^3+3y^2+3y+1.
A puzzle about the two-cube calendar was described in Gardner's column in Scientific American. [1] [2] In the puzzle discussed in Mathematical Circus (1992), two visible faces of one cube have digits 1 and 2 on them, and three visible faces of another cube have digits 3, 4, 5 on them. The cubes are arranged so that their front faces indicate ...
The Rubik's Cube is constructed by labeling each of the 48 non-center facets with the integers 1 to 48. Each configuration of the cube can be represented as a permutation of the labels 1 to 48, depending on the position of each facet. Using this representation, the solved cube is the identity permutation which leaves the cube unchanged, while ...
Mechanically identical to the standard 3×3×3 cube. However, the numbers on the centre pieces force the solver to become aware that each one can be in one of four orientations, thus hugely increasing the total number of combinations. The number of combinations of centre face orientations is 4 6. However, odd combinations (overall odd number of ...
The number of different states that are reachable for cubes of any size can be simply related to the numbers that are applicable to the size 3 and size 4 cubes. Hofstadter in his 1981 paper [ 22 ] provided a full derivation of the number of states for the standard size 3 Rubik's cube.
[1] The maximal number of face turns needed to solve any instance of the Rubik's Cube is 20, [2] and the maximal number of quarter turns is 26. [3] These numbers are also the diameters of the corresponding Cayley graphs of the Rubik's Cube group. In STM (slice turn metric), the minimal number of turns is unknown.
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