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up vote 6 down vote favorite

This is a 3x3 magic square of summation,
in which sums of each row, column, and diagonal are equal.

$$\begin{array}{c|c|c} 4&9&2\\\hline 3&5&7\\\hline 8&1&6 \end{array}$$

Now modify the magic square by defining a simple function $f(x)$,
so it becomes a new magic square of multiplication,
i.e. in which products of each row, column, and diagonal are equal.

$$\begin{array}{c|c|c} f(4)&f(9)&f(2)\\\hline f(3)&f(5)&f(7)\\\hline f(8)&f(1)&f(6) \end{array}$$

Note: Create the function as simple as possible.

share|improve this question
    
Obviously f(x) = 0 X x will work but I guess that's not what we are after. – Radoslav Hristov 1 hour ago
up vote 10 down vote accepted

How about:

$f(x)=c_1\times({c_2}^x)$ with any $c_1$ and $c_2$.

Check:

$f(x)\times f(y)\times f(z)=(c_1\times({c_2}^x))\times(c_1\times({c_2}^y))\times(c_1\times({c_2}^z))={c_1}^3\times{c_2}^{x+y+z}$
As previously $x+y+z$ was equal in the desired columns, rows and diagonals, these products will be as well.

An answer which is not of this form is:

$f(x)=(x+1)\mod2$
As each row, column and diagonal has an odd number, this product will always give 0, although $f$ is $1$ for the even numbers.

share|improve this answer
    
Very elegant solution. – Radoslav Hristov 1 hour ago
    
Thanks, @RadoslavHristov. – elias 1 hour ago
up vote 8 down vote

My Shot:

$f(x) = e^x$.

Reasoning.

$e^x \times e^y \times e^x = e^{x+y+z}$
Since the sums match in the magic square, the products will match in this case. They will all be $e^{15}$.

Second try.

$f(x) = x^0$.
all the products will be 1.

And maybe the simplest function

$f(x) = c$ where c is any number, real, complex, integer, rational, irational.

share|improve this answer
    
Nice, though $x^0$ is just a fancy way to say $1$. Constant $1$ and $e^x$ are both special cases of my generic answer. – elias 1 hour ago
1  
@elias. True. That's why +1'd your answer. I'm a simple man, I think in simple terms. – Marius 1 hour ago
2  
Cool. Your answer of $c$ even beats mine in simplicity, which was asked for in the question. – elias 1 hour ago

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