Ken's POTW

A Magic Cube
In a 3x3x3 cube, there are 27 unit cubes. Is it possible to put the numbers 1 through 27 into the cubes such that the sum in any line of three cubes is the same? If so, show how.

If not,

Source: Original, with influence from many puzzle sources.

Received a solution to the second part from Kirk Bresniker:

> 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.]
On 8/19/97, Craig Gentry sent the following proof that such a cube can't exist:
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

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