The First Solution of the Classical Eulerian Magic Cube Problem of Order Ten
Joseph Arkin · The Fibonacci Quarterly · 1973
In this paper for the first time three Latin cubes of the tenth order have been superimposed to form an Eulerian cube. A Latin cube of the tenth order is defined as a cube of 1000 cells (in ten rows, ten columns, and ten files) in which 1000 numbers consisting of 100 zeros, 100 ones, • • • , 100 nines, are arranged in the cells so that the ten numbers in each row, each column, and each file are different. In this paper, we actually solved two problems, since in addition to having solved the Eulerian cube of order ten, we have also made the cube magic (for the first time). A magic cube is such that the ten cells in each diagonal (or "diameter") and in every row, every file, and every column is the same — namely, 4995 (see [l]). In what follows, it will be noted that each of the ten SQUARES contain 100 cells and each cell contains a three-digit number. Now, if we delete the third digit on the right side in each and every cell, it is easily verified that each of the ten SQUARES has become pairwise orthogonal. In 1779, Euler conjectured that no pair of orthogonal squares exist for n = 2 (mod 4). Then in 1959, the Euler conjecture was shown to be incorrect by the remarkable mathematics of Bose, Shrikande and Parker [2]. Recently (in 1972) Hoggatt and this author extended Bose, Shrikande and Parker 1 s work by finding a way to make the 10 X10 square pairwise o r-thogonal as well as magic. For a square to be magic, each of the two diagonals must have the same sum as in every row and in every column — namely (since we are considering the sum of ten cells with two digits in each cell), 495 (see [3]). Let us label the cells in each square as follows: (row, column, square number) = (r, c, s) = some number in a cell. For example, the number 763 in Square Number 0 reads 763 = (0,0,0), or say we wish to consider the number 338 in Square Number 1: we then write 338 = (6, 2, 1).