Source: Original, with influence from many puzzle sources.
From: http://his.kanazawa-it.ac.jp/~poyo/magic/ > Equi-Sum=42 > > (0) (1) (2) > > 10 23 9 24 7 11 8 12 22 > 26 3 13 1 14 27 15 25 2 > 6 16 20 17 21 4 19 5 18 > > The solution to this took me about two minutes, ten seconds to > type "+magic +cube" in Alta-Vista and the rest of the time to > search the results. Work smart, not hard![This solution has equi-sums through the center cube, but not for the face diagonals.]
A perfectly magic cube is impossible. Since any line of 3 cubes must have a sum of 42, each line must have 0 or 2 odd numbers. Accordingly, each square of nine cubes must have 0, 4, or 6 odd numbers, because it is impossible to arrange 2 or 8 odd numbers so that each row and column of the square has an even number of them. Choose an axis going through the centers of two opposite faces of the cube. Since there are a total of 14 odd numbers, 2 of the 3 squares perpendicular to the axis must have 4 odd numbers, while the other has 6. In each of the two squares having only 4 odds, the odds must all be in corners of the square to ensure that each row, column and diagonal of the square has an even number of them. In the square having 6 odds, the only acceptable configuration is where the 3 evens are on a main diagonal of the square. But now we have a problem, because several of the lines parallel to our axis must have an odd number of odds. Specifically, every line parallel to our axis that goes through an odd number in the 6-odd-square has an odd number of odds. -Craig Gentry