Max nb of Multiplicative Magic Cubes.
by Michael Quist, July 2008
The comment on this problem (find a multiplicative magic cube with max entry < 364) states that the order is open. In fact, it's possible to show that the order must be 4 (assuming you're correct that no 3x3x3 cube can have max entry < 400, which I didn't check).
For order 5, the product of all entries in the cube must be a factor of the product of all *possible* entries, i.e., a factor of 363!
The multiplicative constant must be a factor of the largest x such that x^25 divides 363!, which is x = 2^14 * 3^7 * 5^3 * 7^2 * 11 * 13.
This x has 133 factors in 1...363, so there are enough (barely) to
fill the required 5^3=125 spaces. But now we have restricted the possible entries to these 133 factors, so we can iterate. The new
(smaller) product of all possible entries is 2^221 * 3^120 * 5^66 * 7^44 * 11^26 *
This limits the multiplicative constant to being a factor of 2^8 * 3^4 * 5^2 * 7 * 11.
For order 6 or higher, the proof is easier. Then the multiplicative constant must be a factor of the largest x such that x^(N^2) divides 363!, which is x = 2^9 * 3^4 * 5^2 * 7 * 11 = 79833600 for N=6. This has only 102 factors in 1...363, already less than the required 6^3=216, so the 6x6x6 cube isn't possible either (much less any N>6).
That leaves the 4x4x4. The method can used to give a lower bound for the order 4 cube as well. I haven't done that either, but the lower bound for the max entry is somewhere between 200 and 240.
Here are the lower bounds provided by the method I described:
Slightly stronger lower bounds are provided by determining by looking at the largest possible multiplicative constant derived from the allowed entries, and comparing it to the product of the smallest N^3 allowed entries. If the latter is larger than the former, then the proposed max entry is invalidated. For instance, there are exactly 64 allowed entries if we set max entry = 217 for order 4. Their product is larger than the largest 16th power that divides their product (i.e., their product is not a perfect 16th power), so max entry = 217 is invalid. The improved bounds are:
As you can see, the upper bounds provided by your table are generally just a few times larger than these (roughly N^4.25), except for N=3, N=6, and N=10, which are unusually high.
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