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Is it possible to form a $3\mbox{x}3$ grid containing the set of numbers: $${1,2,4,8,16,32,64,128,256}$$

in such a way that the product of the numbers in every row, column and diagonal are the same? If so, how?

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  • $\begingroup$ This is just a magic square with powers of two? $\endgroup$ – xnor Nov 6 '14 at 0:32
  • $\begingroup$ Why did this get downvoted? $\endgroup$ – warspyking Nov 6 '14 at 0:35
  • $\begingroup$ @xnor Well a magic square deals with the sum (result of an addition equation) this question deals with the product. $\endgroup$ – warspyking Nov 6 '14 at 0:40
  • $\begingroup$ Hmm. Does it count as a duplicate if the answer has to explain why it's basically exactly the same? Probably not. $\endgroup$ – TheRubberDuck Nov 6 '14 at 2:49
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This is just a standard Lo Shu Square with exponents and multiplication instead of addition. But because the multiplication of exponents with the same base is analogous to the addition of their exponents, the solution here is just the same as $2^\left(L-1\right)$, where $L$ is the standard Lo Shu Square.

As an expanded explanation, your numbers are better expressed as: $2^0, 2^1, 2^2, 2^3, 2^4, 2^5, 2^6, 2^7, 2^8$. We can replace the whole "product" idea completely with just adding the powers of these exponents, by the rules of exponentiation. $a^n*a^m=a^\left(n+m\right)$, so we need to make a magic square that adds the numbers $[0,7]$.

We know that the numbers span $[0,8]$, and since we know the regular Lo Shu Square operates from $[1,9]$, we just subtract one from every position, making each row and column add to $15-3=12$.

$$\begin{array}{|c|c|c|}\hline 4&9&2\\\hline 3&5&7\\\hline 8&1&6\\\hline\end{array} \overset{-1}{\rightarrow} \begin{array}{|c|c|c|}\hline 3&8&1\\\hline 2&4&6\\\hline 7&0&5\\\hline\end{array}$$

If we exponentiate two by this grid we end up with a grid such that all numbers, when multiplied, will be the same (by rules of exponents). If we take any three numbers $a+b+c=12$ from one row or column, then $2^a*2^b*2^c=2^{(a+b+c)}=2^{12}$. We can see that this is evidently true:

$$\begin{array}{|c|c|c|}\hline 2^3&2^8&2^1\\\hline 2^2&2^4&2^6\\\hline 2^7&2^0&2^5\\\hline\end{array} = \begin{array}{|c|c|c|}\hline 8&256&2\\\hline 4&16&64\\\hline 128&1&32\\\hline\end{array}$$

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