-5
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Using consecutive odd numbers from negative five to eleven, make a 3x3 magic square

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7
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Take the standard $3 \times 3$ square, double each entry and subtract $7$. That is the linear transformation that takes $1$ to $-5$ and $9$ to $11$

$$\begin {array}{c c c} 9&-5&5\\-1&3&7\\1&11&-3 \end {array}$$

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1
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A less-mathematical way of looking at this, that might be easier to intuitively understand:

Write out the numbers you want to put into your square in order, and then map them to the integers $1$ through $9$, which are generally used for the simplest $3\times 3$ magic square:

$$\begin{array}{rrrrrrrrr} -5 & -3 & -1 & 1 & 3 & 5 & 7 & 9 & 11\\ \color{red}1 & \color{red}2 & \color{red}3 & \color{red}4 & \color{red}5 & \color{red}6 & \color{red}7 & \color{red}8 & \color{red}9\\ \end{array} $$

As long as your list of numbers is evenly-spaced (the difference between any two consecutive numbers is the same), this strategy will work.

Then, take your standard magic square:

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

and substitute in the mapped values:

$$\begin{array}{c|c|c} 9 & -5 & 5\\ \hline -1 & 3 & 7\\ \hline 1 & 11 & -3\\ \end{array}$$

The result should be the magic square you seek.

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