# Productive Squares

Consider a productive square of size $$n$$ to be an $$n\times n$$ grid filled with a permutation of the integers in $$[1, n^2]$$, such that the product of all the numbers along the first row is equal to the product of all the numbers along the first column, and the same applys to the second row/column, the third row/column, etc. We already know that 3~6 are solvable, and anything above $$10$$ is unsolvable (and $$9$$ is unsolvable as well).

This got me interested: is there a configuration for $$7/8/10$$?

• For the 10x10 square, I know the numbers on the diagonal. But that's as far as I can reason it, before having to brute-force it. Feb 4 at 13:10

• @RobPratt my MIP isn't terminating for $N=7$, its running for 6 minutes already... is it normal? Feb 5 at 9:37
• And how exactly should I implement the $\log$ indicated in the answer, since I'm a little worried about rounding issues? Feb 5 at 9:39
• (It works fine for $N=1,2,3,4,5,6$) $N=6$ takes about 1 second. Feb 5 at 9:40