This week’s El Pais math challenge was easier to solve than the previous ones (maybe that’s why I managed to solve it without using the computer this time 😄).
The problem is the following: find a multiplicative magic square, that is, a 3x3 square in which the product of the numbers in every row, column, and diagonal is the same (unknown) value. All numbers are distinct positive integers. It is known that the number in the center cell is 15.
My approach was to assign a variable name to each cell:
$$ \large{ \begin{array}{|c|c|c|} \hline a & b & c \\\hline d & 15 & f \\\hline g & h & i \\\hline \end{array}} $$This way I obtained equations such as:
$$ \begin{align} a * b * c & = a * d * g \\ a * b * c & = a * 15 * i \\ a * b * c & = d * 15 * f \\ a * d * g & = g * h * i \\ ... & \end{align} $$which allows some terms to cancel out and, after several manipulations, I arrived at the following magic square (up to rotation, since rotations are equivalent):
$$ \large{ \begin{array}{|c|c|c|} \hline \frac{15^3}{b·c} & b & c \\\hline \frac{b·c^2}{15^2} & 15 & \frac{15^4}{b·c^2} \\\hline \frac{15^2}{c} & \frac{15^2}{b} & \frac{b·c}{15} \\\hline \end{array}} $$Now we can see that the product of any row, column, or diagonal gives the same result (as required), and that this result is $15^3 = 3375$. This was the unknown product. Furthermore, since the cells can only contain positive integers, the number of possible solutions is finite (in fact, unique if rotations are excluded).
We can also see that both $b$ and $c$ must be factors of $15^2$, or in other words, both $b$ and $c$ must be one of: $1, 3, 5, 3^2, 3 * 5$ (already used in the center), $5^2, 3^2 * 5, ...$
That is, every possible combination of the form $3^x * 5^y$ (with both $x$ and $y$ taking values in $0, 1, 2, 3$). There are 16 possible combinations. Let us start by assuming $b = 1$; then the resulting square is:
$$ \large{ \begin{array}{|c|c|c|} \hline \frac{15^3}{c} & 1 & c \\\hline \frac{c^2}{15^2} & 15 & \frac{15^4}{c^2} \\\hline \frac{15^2}{c} & 15^2 & \frac{c}{15} \\\hline \end{array}} $$The bottom-right cell reveals that 15 must be a factor of $c$ (and $c$ cannot be 15, because the value in the upper-right cell would then be the same as the one in the center).
The bottom-left cell also tells us that $c$ must be a factor of $15^2$ (which is $3^2 * 5^2$). This means that $c$ must be a product of either 3 or 9 with either 5 or 25 (only 3 combinations are possible, since $3 * 5 = 15$ has already been discarded).
I tried $9 * 5 = 45$ and... eureka! $c = 45$. Replacing $c$ with its value gives the following magic square:
$$ \large{ \begin{array}{|c|c|c|} \hline 75 & 1 & 45 \\\hline 9 & 15 & 25 \\\hline 5 & 225 & 3 \\\hline \end{array}} $$