Multiplicative magic cubes

What are multiplicative magic cubes?

Multiplicative magic cubes are cubes which are magic using multiplication instead of addition. The used numbers can't be consecutive, but they are required to be distinct. A reminder of some definitions, similar to additive cubes:

It seems that the first multiplicative magic cubes were published in 1913 by Harry A. Sayles in The Monist in 1913. This paper was republished in the book Magic squares and cubes by W.S. Andrews, where we can find (page 293) two cubes:

In 2002, Marián Trenkler, Slovakia, published a paper on multiplicative magic squares and cubes in Obzory Matematiky, Fyziky a Informatiky 1/2002 (31), pages 9-16, with:

Is it possible to do better? Cube of same orders, but using smaller products P, or smaller Max nb? Yes, both for the orders 4 and 5!

After having studied multiplicative magic squares, here are my best results on cubes, 3rd-order to 11th-order, done in January 2006 (and May 2006 for 6th and 10th-orders, and June 2007 for a better 4th-order cube). My goal is always to minimize the magic product and the max nb.

Many thanks to Edwin Clark, Mathematics Department of the University of South Florida, USA, for his checkings of all my multiplicative cubes, confirming that they have all the announced properties.

3rd-order multiplicative magic cubes

It is easy to prove that any 3rd-order multiplicative magic cube must have P = (center)^3. With the two different possible constructions using the integer 1

we can produce exactly 4 different 3rd-order multiplicative magic cubes with the same P = (2·3·5)^3 = 27000: one cube from the first construction, and three from the second construction. They use the same set of integers, but they are "different" because from any of these 4 cubes, it is impossible to produce any other of them, using symmetries and rotations.

It is impossible to construct better 3rd-order cubes with smaller P. But, using Min nb > 1, it is possible to construct 3rd-order cubes with smaller Max nb (and bigger P), the smallest possible Max nb being 400. Here are some examples with Max nb < 900:

And it is possible to construct 3rd-order semi-magic cubes using smaller constants. In the three examples below, all the rows, columns and pillars have the same magic product P = 7560, but some of their 4 triagonals do not have the same product: the example on the right is a very nearly magic cube, only one triagonal is incorrect!!!

And what about adding magic plane diagonals? Both for additive and multiplicative magic cubes, it is impossible to construct 3rd-order perfect magic cubes.

4th-order multiplicative magic cubes

The Sayles's cube and the Trenkler's cube have the same characteristics: P = 57153600, Max nb = 7560. Is it possible to construct 4th-order cubes with smaller constants? Yes! My best cubes have a P more than 8 times smaller, and a Max nb more than 20 times smaller:

Best known 4th-order multiplicative magic cubes
P = 6,486,480 and Max nb = 546 (left cube, January 2006),
P = 17,297,280 and Max nb = 364 (right cube, June 2007)
The right cube is the best known cube -of any order- using the smallest possible integers
(click on the image to enlarge a cube)

They are my two best cubes, but I am not sure I have found the best possible cubes. Who will be able to construct better 4th-order cubes (with smaller P or smaller Max nb)?  Is it possible to construct a multiplicative magic cube (of any order!) using integers < 364? Of any order, because we can remark that the Max nb = 364 of the 4th-order cube on the right is smaller than the Max nb = 400 used for the best possible 3rd-order cubes.
In July 2008, Michael Quist worked on this enigma #5, and found that any 4th-order cube must have Max nb ≥ 221 and that any 5th-order (or above) cube must have Max nb ≥ 442. If we suppose that his work is correct (read it here), and if we suppose that any 3rd-order cube must have Max nb ≥ 400, then the enigma is limited to the 4th-order, and is equivalent to: is it possible to construct a multiplicative magic cube, so of 4th order, and having 221 ≤ Max nb < 364?

Some plane diagonals of the two above cubes are magic, but not all of them: both for additive and multiplicative magic cubes, it is impossible to construct 4th-order perfect magic cubes. The cube on the left uses the following construction with (a, b, c, d, e, f) = (2, 3, 5, 7, 11, 13):

In January 2010, Max Alekseyev, Dept of Computer Science & Engineering, University of South Carolina, found another 4th-order magic cube having the same Max nb = 364 than my above cube, but with a smaller P. None of its 24 small diagonals are magic (8 small diagonals are magic in my cube), but this is not a problem because not needed in a magic cube: only rows + columns + pillars + 4 triagonals have to be magic. An excellent cube constructed by Max!

5th-order multiplicative magic cubes

The Trenkler's cube published in 2002 has P = 35286451200, Max nb = 2448. Is it possible to construct 5th-order cubes with smaller constants? Yes! My best cube has a P more than 2 times smaller, and a Max nb more than 2 times smaller:

Best known 5th-order multiplicative magic cube,
P = 16,761,064,320 and Max nb = 1026
(click on the image to enlarge it)
 

All its rows, columns, pillars and 4 triagonals are magic. And this cube has a very nice additional property: all its broken triagonals are magic, as the example in blue.

But unfortunately the plane diagonals are not magic: it is not a perfect magic cube. However, it is possible to construct 5th-order perfect magic cubes: use the first known 5th-order (additive) perfect magic cube constructed by Walter Trump and me in 2003, and replace each number n by 2^(n-1). Then you will get a multiplicative perfect magic cube... but tedious... using very big numbers: Max nb = 2^(125-1) = 2.13 · 10^37, and P = 2^310 = 2.09 · 10^93.

6th-order multiplicative magic cubes

I have constructed two cubes:

These cubes are only semi-magic: their 4 triagonals are not magic.

Later, in May 2006, I constructed a magic cube, this time with 4 magic triagonals. This added feature has a cost, bigger P and Max nb than the previous semi-magic cubes:

It is possible to construct 6th-order perfect magic cubes: use the first known 6th-order (additive) perfect magic cube constructed by Walter Trump in 2003, and replace each number n by 2^(n-1). Then you will get a multiplicative perfect magic cube... but tedious... using very big numbers: Max nb = 2^(216-1) = 5.27 · 10^64, and P = 2^645 = 1.46 · 10^194.

7th-order multiplicative magic cubes

My two best magic cubes are:

As the above 5th-order magic cube, all their broken triagonals and 4 entire triagonals are magic, but their plane diagonals are not magic: they are not perfect magic cubes. With the same P and Max nbs, I have successfully constructed cubes with all their magic plane diagonals... but unfortunately loosing 2 magic triagonals on the 4 triagonals.

Keeping magic plane diagonals, I have constructed two cubes with 3 magic triagonals (now, only one triagonal is bad!) with the cost to use bigger characteristics:

and finally succeeded in constructing perfect magic cubes with 4 magic triagonals, but with again bigger characteristics. The cost to get the 4th triagonal is very expensive!

But, even with these big characteristics, the cube P = 1.41 · 10^27 is the best known perfect magic cube -of any order- using the smallest magic product. Who will succeed constructing a perfect cube -of any order (5, 6, 7,...)- having a smaller magic product?

Perfect cubes using smaller integers are known: see my 11th-order pandiagonal perfect cube with Max nb = 24,992.

8th to 11th-order multiplicative magic cubes

See the summary in the table at the beginning of this page. And see the page on the pandiagonal perfect multiplicative magic cubes.

In the honour of January 2006 when the cubes were created, the two first numbers used in the 10th and 11th-order cubes are: "2006" and "1"!

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