A 3×3 square is filled out with 9 positive integers such that the product of each row, column, and diagonals are equal. The sum of all 4 corners is less that 10. Find all possible configurations
In this case, a square cannot be rotated to make another one. You can repeat numbers too.
I thought of this as a systematic question, carefully writing down all possibilities, but that took way too long. Is there a faster way?
[Math] Magic square but with multiplication
magic square
Best Answer
Every magic multiplication 3x3 square is of the form: $$\begin{bmatrix} \frac{a}{b} & abc & \frac{a}{c} \\ \frac{ab}{c} & a & \frac{ac}{b} \\ ac & \frac{a}{bc} & ab \end{bmatrix}$$ with the magic product $a^3$
If we assume all entries are integers, the square is rewritten as $$\begin{bmatrix} ac & ab^2c^2 & ab \\ ab^2 & abc & ac^2 \\ abc^2 & a & ab^2c \end{bmatrix}$$ with the magic product $a^3b^3c^3$
The sum of the corners is then $ac+ab+abc^2+ab^2c=a(1+bc)(b+c)<10$. Since $a,b,c$ must be positive integers, it follows that $1\leq a,b,c \leq 2$.
The only valid solutions are $(a,b,c)=(1,1,1),(2,1,1),(1,2,1),(1,1,2)$ corresponding to the squares: $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}, \begin{bmatrix} 1 & 4 & 2 \\ 4 & 2 & 1 \\ 2 & 1 & 4 \end{bmatrix}, \begin{bmatrix} 2 & 1 & 4 \\ 4 & 2 & 1 \\ 1 & 4 & 2 \end{bmatrix}$$ The last two squares actually just the same square rotated. So there are 3 possible such squares. Definitely not too many to count. $$\begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix}, \begin{bmatrix} 2 & 2 & 2 \\ 2 & 2 & 2 \\ 2 & 2 & 2 \end{bmatrix}, \begin{bmatrix} 1 & 4 & 2 \\ 4 & 2 & 1 \\ 2 & 1 & 4 \end{bmatrix}$$