Multiplicative magic cubes
What are multiplicative magic cubes?
Multiplicative magic cubes are cubes which are magic using multiplication instead of addition. The numbers used can't be consecutive, but they must 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:
1 
90 
300 

150 
4 
45 

180 
75 
2 
60 
25 
18 
9 
30 
100 
50 
36 
15 

450 
12 
5 
20 
225 
6 
3 
10 
900 
In 2002, Marián Trenkler, Slovakia, published a paper on multiplicative magic squares and cubes in Obzory Matematiky, Fyziky a Informatiky 1/2002 (31), pages 916, with:
Is it possible to do better? Cube of the same orders, but using smaller products P, or smaller Max nb? Yes, both for the orders 4 and 5!
After multiplicative magic squares, here are the best known cubes, 3rdorder to 11thorder. The goal is always to minimize the magic product and the max nb.
Order 
Magic product P 
Max nb 
Multiplicative cube 
Comments 
3 
(c) 6 720 
224 
semimagic 

7 560 
135 
semimagic 


(a) 27 000 
900 
magic 
The best known magic cube (of any order) using the smallest P 

216 000 
400 
magic 


4 
(b) 4 324 320 
351 
magic 
The best known magic cubes (of any order) using the smallest Max nb 
5 
(b) 13 967 553 600 
855 
magic 
+ magic broken triagonals 
(b) 101 625 502 003 200 000 
2 976 750 
perfect magic 
The best known perfect magic cube (of any order) using the smallest P 

(b) 104 064 514 051 276 800 000 
250 880 
perfect magic 


6 
(b) 117 327 450 240 000 
5 225 
magic 

(b) 23 959 607 303 503 872 000 
849 420 
perfect magic 


(b) 80 863 674 649 325 568 000 000 
66 924 
perfect magic 


7 
(b) 897 612 484 786 617 600 
3 367 
magic 
+ magic broken triagonals 
(b) 19 407 837 508 899 840 000 
2 912 
magic 
+ magic broken triagonals 

1 411 407 979 783 492 239 360 000 000 
1 259 712 
perfect magic 


5 750 476 043 814 094 602 240 000 000 
862 400 
perfect magic 


8 
58 165 289 014 172 820 480 000 
11 100 
magic 

297 508 272 407 615 683 814 400 
7 520 
magic 


89 518 183 823 250 314 294 722 560 000 
17 297 280 
pandiag. perfect magic (*) 
The best known pandiag. perfect magic cube (of any order) using the smallest P 

9 
1 845 817 281 575 760 285 112 320 000 
19 350 
magic 

25 544 060 268 917 882 612 304 384 000 
12 150 
magic 


265 237 261 271 449 982 022 984 892 416 000 
591 192 
pandiag. perfect magic (*) 


10 
117 218 854 345 145 394 654 241 228 800 000 
22 125 
semimagic 

602 839 822 346 462 029 650 383 462 400 000 
17 400 
semimagic 


~ 3.50 E+44 
810 810 000 
magic 


(b) ~ 2.62 E+51 
1 920 996 000 
perfect magic 


(b) ~ 3.52 E+59 
9 018 009 000 
pandiag. perfect magic (*) 


11 
9 009 441 144 967 875 033 124 980 845 568 000 000 
46 620 
pandiag. perfect magic (*) 

174 930 251 129 029 312 377 859 321 968 844 800 000 
24 992 
pandiag. perfect magic (*) 
The best known perfect magic cube (of any
order), and 
Many thanks to Edwin Clark, Mathematics Department of the University of South Florida, USA, for checking in 2006 all my multiplicative cubes, confirming that they have all the announced properties. 
3rdorder multiplicative magic cubes
It is easy to prove that any 3rdorder multiplicative magic cube must have P = (center)^3. With two different possible constructions using the integer 1
1 
abc² 
a²b²c 

a 
a²bc² 
b²c 
ab²c 
a² 
bc² 
a²b²c 
1 
abc² 

a²bc² 
b²c 
a 
bc² 
ab²c 
a² 




a²bc 
b² 
ac² 
a²bc 
b² 
ac² 

c² 
abc 
a²b² 
c² 
abc 
a²b² 

ab² 
a²c² 
bc 
ab² 
a²c² 
bc 




ab²c² 
a²c 
b 
b²c² 
ac 
a²b 

a²b 
b²c² 
ac 
ab 
a²b²c² 
c 

c 
ab 
a²b²c² 
a²c 
b 
ab²c² 
we can produce exactly 4 different 3rdorder 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 ones, using symmetries and rotations.
1 
150 
180 

2 
300 
45 

3 
450 
20 

5 
450 
12 
90 
4 
75 
180 
1 
150 
180 
1 
150 
300 
1 
90 

300 
45 
2 
75 
90 
4 
50 
60 
9 
18 
60 
25 






60 
9 
50 
60 
9 
50 
90 
4 
75 
150 
4 
45 

25 
30 
36 
25 
30 
36 
25 
30 
36 
9 
30 
100 

18 
100 
15 
18 
100 
15 
12 
225 
10 
20 
225 
6 






450 
20 
3 
225 
10 
12 
100 
15 
18 
36 
15 
50 

12 
225 
10 
6 
900 
5 
6 
900 
5 
10 
900 
3 

5 
6 
900 
20 
3 
450 
45 
2 
300 
75 
2 
180 
It is impossible to construct better 3rdorder cubes with a smaller P. But, using Min nb > 1, it is possible to construct 3rdorder cubes with smaller Max nb (and bigger P), the smallest possible Max nb being 400. Here are some examples with Max nb < 900:
90 
80 
30 

320 
150 
36 

200 
192 
45 
240 
9 
100 
180 
32 
300 
288 
25 
240 

10 
300 
72 
30 
360 
160 
30 
360 
160 





48 
225 
20 
60 
288 
100 
96 
225 
80 

25 
60 
144 
200 
120 
72 
100 
120 
144 

180 
16 
75 
144 
50 
240 
180 
64 
150 





50 
12 
360 
90 
40 
480 
90 
40 
480 

36 
400 
15 
48 
450 
80 
60 
576 
50 

120 
45 
40 
400 
96 
45 
320 
75 
72 
And it is possible to construct 3rdorder semimagic 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!!!
1 
56 
135 

1 
42 
180 

1 
30 
252 
72 
15 
7 
54 
20 
7 
90 
28 
3 

105 
9 
8 
140 
9 
6 
84 
9 
10 





84 
45 
2 
84 
45 
2 
60 
63 
2 

5 
14 
108 
5 
14 
108 
7 
6 
180 

18 
12 
35 
18 
12 
35 
18 
20 
21 





90 
3 
28 
90 
4 
21 
126 
4 
15 

21 
36 
10 
28 
27 
10 
12 
45 
14 

4 
70 
27 
3 
70 
36 
5 
42 
36 
In February 2013, André LFS Bacci, Brasilia, Brazil, constructed this semimagic cube with a smaller P:
21 
20 
16 
32 
42 
5 
10 
8 
84 


2 
56 
60 
15 
4 
112 
224 
30 
1 


160 
6 
7 
14 
40 
12 
3 
28 
80 
And what about adding magic plane diagonals? Both for additive and multiplicative magic cubes, it is impossible to construct 3rdorder perfect magic cubes.
4thorder multiplicative magic cubes
Sayles's cube and Trenkler's cube have the same characteristics: P = 57153600, Max nb = 7560. Is it possible to construct 4thorder 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:
4thorder multiplicative magic cubes, by Christian Boyer
P = 6,486,480
and Max nb = 546 (left cube, January 2006),
P = 17,297,280 and Max nb = 364
(right cube, June 2007)
(click
on the image to enlarge a cube)
Some plane diagonals of my two cubes are magic, but not all of them: both for additive and multiplicative magic cubes, it is impossible to construct 4thorder perfect magic cubes.
These were my two best cubes, but not sure I found the best possible cubes. That's why I asked these next questions. Who will be able to construct better 4thorder 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 notice that the Max nb = 364 of the 4thorder cube on the right is smaller than the Max nb = 400 used for the best possible 3rdorder cube.
In July 2008, Michael Quist worked on this problem, and found that any 4thorder cube must have Max nb ≥ 221 and that any 5thorder (or higher) cube must have Max nb ≥ 442. If we suppose that his work is correct (read it here), and if we suppose that any 3rdorder cube must have Max nb ≥ 400, then the enigma is limited to the 4thorder, and is equivalent to: is it possible to construct a multiplicative magic cube, of 4th order, and having 221 ≤ Max nb < 364?
In January 2010, Max Alekseyev, Dept of Computer Science & Engineering, University of South Carolina, found another 4thorder magic cube having the same Max nb = 364 as 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!
1 
110 
224 
351 
130 
8 
297 
28 
308 
27 
13 
80 
216 
364 
10 
11 


231 
12 
78 
40 
96 
273 
5 
66 
6 
55 
168 
156 
65 
48 
132 
21 


144 
91 
15 
44 
77 
18 
52 
120 
195 
32 
198 
7 
4 
165 
56 
234 


260 
72 
33 
14 
9 
220 
112 
39 
24 
182 
20 
99 
154 
3 
117 
160 
Toshiro Shirakawa in 2010 (Kuwana, Japan, 1983  )
And finally in April 2010, Toshihiro Shirakawa solved my enigma #5 with this excellent cube using smaller integers: Max nb = 351 < 364. The magic product is also smaller than the above cubes. Congratulations!!! Toshihiro lives in Ebina Kanagawa, Japan. He is a programmer, and does mathematics as a hobby.
April 2010. Best known 4thorder multiplicative magic cube,
by Toshihiro Shirakawa
and best known cube of any order using the smallest possible integers
P = 4,324,320
and Max nb = 351
(click
on the image to enlarge it)
His method is easy and ingenious, without any computing. He used, directly from this website, my own 4x4 multiplicative semimagic square having the smallest possible product that I constructed in 2005, with magic product 4320 = 2^{5} * 3^{3} * 5 and Max nb 27:
16 
1 
10 
27 
= 
1A 
1B 
1C 
1D 
5 
24 
18 
2 
2A 
2B 
2C 
2D 

6 
12 
3 
20 
3A 
3B 
3C 
3D 

9 
15 
8 
4 
4A 
4B 
4C 
4D 
Noticing that this square used only factors 2, 3 and 5, he cleverly reused it, combined with the four factors (E, F, G, H) = (1, 11, 7, 13). That's why his magic product is 4320(from the square) *7*11*13(his factors) = 4324320, and his Max nb is 27(from the square) * 13(his max factor) = 351. And 351 is less than 364, as asked in the enigma. That's it! Er... shame on me... to have been unable to think to this ingenious method... using my own square...
1A 
1B 
1C 
1D 
* 
E 
F 
G 
H 
1B 
1A 
1D 
1C 
H 
G 
F 
E 

1C 
1D 
1A 
1B 
F 
E 
H 
G 

1D 
1C 
1B 
1A 
G 
H 
E 
F 




2A 
2B 
2C 
2D 
F 
E 
H 
G 

2B 
2A 
2D 
2C 
G 
H 
E 
F 

2C 
2D 
2A 
2B 
E 
F 
G 
H 

2D 
2C 
2B 
2A 
H 
G 
F 
E 




3A 
3B 
3C 
3D 
G 
H 
E 
F 

3B 
3A 
3D 
3C 
F 
E 
H 
G 

3C 
3D 
3A 
3B 
H 
G 
F 
E 

3D 
3C 
3B 
3A 
E 
F 
G 
H 




4A 
4B 
4C 
4D 
H 
G 
F 
E 

4B 
4A 
4D 
4C 
E 
F 
G 
H 

4C 
4D 
4A 
4B 
G 
H 
E 
F 

4D 
4C 
4B 
4A 
F 
E 
H 
G 
I can now give the method used for my cube (P, MaxNb) = (17297280, 364) which was challenged with enigma #5. This cube was constructed from the Eulerian (or GraecoLatin) cube below, using the best possible set (A, B, C, D) (I, J, K, L) (a, b, c, d) generating 64 distinct integers and generating the smallest MaxNb:
D 
C 
A 
B 
* 
I 
L 
J 
K 
* 
d 
b 
a 
c 
B 
A 
C 
D 
K 
J 
L 
I 
a 
c 
d 
b 

C 
D 
B 
A 
L 
I 
K 
J 
c 
a 
b 
d 

A 
B 
D 
C 
J 
K 
I 
L 
b 
d 
c 
a 





B 
A 
C 
D 
L 
I 
K 
J 
a 
c 
d 
b 

D 
C 
A 
B 
J 
K 
I 
L 
d 
b 
a 
c 

A 
B 
D 
C 
I 
L 
J 
K 
b 
d 
c 
a 

C 
D 
B 
A 
K 
J 
L 
I 
c 
a 
b 
d 





C 
D 
B 
A 
J 
K 
I 
L 
c 
a 
b 
d 

A 
B 
D 
C 
L 
I 
K 
J 
b 
d 
c 
a 

D 
C 
A 
B 
K 
J 
L 
I 
d 
b 
a 
c 

B 
A 
C 
D 
I 
L 
J 
K 
a 
c 
d 
b 





A 
B 
D 
C 
K 
J 
L 
I 
b 
d 
c 
a 

C 
D 
B 
A 
I 
L 
J 
K 
c 
a 
b 
d 

B 
A 
C 
D 
J 
K 
I 
L 
a 
c 
d 
b 

D 
C 
A 
B 
L 
I 
K 
J 
d 
b 
a 
c 
My other cube (P, MaxNb) = (6486480, 546) used a similar Eulerian cube, but with the best possible set generating 64 distinct integers and generating the smallest product: (1, 2, 3, 6), (1, 4, 5, 7), (1, 9, 11, 13).
5thorder multiplicative magic cubes
Trenkler's cube published in 2002 has P = 35286451200, Max nb = 2448. Is it possible to construct 5thorder 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:
January 2006: 5thorder multiplicative magic cube, by Christian
Boyer
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.
In May 2010, Toshihiro Shirakawa constructed a 5thorder multiplicative magic cube with smaller characteristics than my above cube: P = 13,967,553,600 and MaxNb = 855. Also with magic broken triagonals. This is today the best known 5thorder magic cube! You can see it in the downloadable Excel file, below the table. Similarly to his 4thorder method, he used my 5thorder multiplicative magic square (reproduced below), combined with (1, 11, 13, 17, 19). My above cube was an Eulerian cube (1, 2, 3, 4, 6) (1, 5, 7, 8, 9) (1, 11, 13, 17, 19).
12 
35 
1 
40 
18 
36 
2 
24 
7 
25 
14 
45 
15 
4 
8 
5 
16 
42 
30 
3 
10 
6 
20 
9 
28 
Unfortunately the plane diagonals of our two cubes are not magic: they are not perfect magic cubes. However, it is possible to construct 5thorder perfect magic cubes: use the first known 5thorder (additive) perfect magic cube constructed by Walter Trump and me in 2003, and replace each number n by 2^(n1). Then you will get a multiplicative perfect magic cube... but tedious... using very big numbers: Max nb = 2^(1251) = 2.13 · 10^37, and P = 2^310 = 2.09 · 10^93. In December 2012, Toshihiro Shirakawa constructed two far better 5thorder perfect magic cubes: one with Max nb = 250 880, one with P = 2520^5 = 1.01 · 10^17. Astonishingly, but with a larger order, perfect cubes using smaller integers are known: see my 11thorder pandiagonal perfect cube with Max nb = 24 992 < 250 880.
6thorder multiplicative magic cubes
In January 2006, I constructed two cubes:
These cubes are only semimagic: 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 semimagic cubes:
Then in June 2010, Toshihiro Shirakawa constructed a magic cube with much smaller characteristics than my poor above cube:
This is today the best known 6thorder magic cube!
It is possible to construct 6thorder perfect magic cubes: use the first known 6thorder (additive) perfect magic cube constructed by Walter Trump in 2003, and replace each number n by 2^(n1). Then you will get a multiplicative perfect magic cube... but tedious... using very big numbers: Max nb = 2^(2161) = 5.27 · 10^64, and P = 2^645 = 1.46 · 10^194. In December 2012 and February 2013, Toshihiro Shirakawa constructed two far better 6thorder perfect magic cubes: one with Max nb = 66 924, one with P = 2 882 880^3 = 2.39 · 10^19. Astonishingly, but with a larger order, perfect cubes using smaller integers are known: see my 11thorder pandiagonal perfect cube with Max nb = 24 992 < 66 924.
7thorder multiplicative magic cubes
In January 2006, my two best magic cubes were:
But in May 2010, Toshihiro Shirakawa constructed better cubes:
As the above 5thorder magic cubes, 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 still in January 2006 two cubes with 3 magic triagonals (now, only one triagonal is bad!) with the cost of using bigger characteristics:
and finally succeeded in constructing perfect magic cubes with 4 magic triagonals, but with again bigger characteristics. To get the 4th triagonal is very expensive!
8th to 11thorder 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.
To celebrate January 2006 when the cubes were created, the two first numbers used in my 10th and 11thorder cubes are: "2006" and "1"!
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