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On a three by three checkerboard, the $9$ numbers $1, 3, 9, 27, 81, ..., 6561$ are placed so that the product of the $3$ numbers in any row or column is the same. What is that product as an integer?

So $1, 3, 9, 27, 81, ..., 6561$ correspond to $3^0, 3^1, 3^2, 3^3, 3^4,..., 3^8$. Since multiplying powers of 3 means adding their exponents, this becomes the same as a problem asking to arrange $0$ to $8$ so that their sums in rows, columns and diagonals are all the same (at least for the first part). For symmetry, I put $3^4$ in the middle of the $3 \times 3$ grid. And after a few trials, I was able to get a pattern that works — the exponents must sum to $12$.

However, I was wondering, if given any problems like this (variations of magic squares, I think), can we figure out the sum of the entries in each row/column/diagonal directly without having to do much trial and error?

space
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2 Answers2

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$\color{red}{\text{Here}}$ I point out this:

Every order-3 magic square can be rotated and reflected so that the smallest element is in the center of the left column and the next-smallest element in the bottom of the right column. Having made this transformation, and ordered the elements from 1 to 9, there are two possible outcomes for the 'arrays of orders'.

$$\text{Type 1: } \; \begin{bmatrix} 8 & 3 & 4\\ 1 & 5 & 9 \\ 6 & 7 & 2 \\ \end{bmatrix} \qquad \text{Type 0: } \; \begin{bmatrix} 8 & 4 & 3\\ 1 & 5 & 9 \\ 7 & 6 & 2 \\ \end{bmatrix}$$

The Type $1$ array is, in fact a magic square itself. However, the Type $2$ array is not.

In your case, you want to, first, take the Type $1$ array and subtract $1$ from each element.

\begin{bmatrix} 7 & 2 & 3\\ 0 & 4 & 8 \\ 5 & 6 & 1 \\ \end{bmatrix}

Note that each row, column and diagonal sums to $12$. Next create a new array by raising $3$ to the corresponding element in each cell.

\begin{bmatrix} 3^7 & 3^2 & 3^3\\ 3^0 & 3^4 & 3^8 \\ 3^5 & 3^6 & 3^1 \\ \end{bmatrix}

Accordingly, the product of the elements of each row, column and diagonal will be $3^{12}$.

Steven Alexis Gregory
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In your problem you have $0+1+2+3+...+ 8 =\frac{8\times 9}{2} =36$ divided into $3$ equal parts so each part is $12$.

In general you have $0+1+2+3+...+(n^2-1) =\frac{(n^2-1)\times (n^2)}{2} $ divided into $n$ equal parts and you get $\frac {n(n^2-1)}{2}$

Note that for $n=3$ the two answers match.

Mohammad Riazi-Kermani
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