Open Contests

The number $36$ has $9$ positive divisors $1$, $2$, $3$, $4$, $6$, $9$, $12$, $18$, $36$. Let the first row be $18$, $1$, $12$, second row $4$, $6$, $9$, and third row $3$, $36$, $2$. Then the product of each row, column and diagonal is $216$.

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Alternative solution.

For each prime $p$ the number $p^8$ has exactly $9$ divisors $p^0,p^1,\dots ,p^8$. It is known that the numbers $1$ to $9$ can be placed as a $3\times 3$ magic square, with an equal sum of the numbers in each row, column and diagonal. By subtracting $1$ from each number, we reduce the sum of each row, column and diagonal by $3$. By replacing in the magic square each number $i$ by the respective power $p^i$ we obtain a placement of the divisors of $p^8$ in which each row, column and diagonal has an equal product.