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 March 29th, 2012, 10:36 AM #1 Senior Member   Joined: Feb 2012 Posts: 623 Thanks: 1 7x7 magic square Just for fun, I made a magic square: $\begin{bmatrix} 8 & 16 & 24 & 32 & 40 & 48 & 56 \\ 31 & 39 & 47 & 55 & 14 & 15 & 23 \\ 54 & 13 & 21 & 22 & 30 & 38 & 46 \\ 28 & 29 & 37 & 45 & 53 & 12 & 20 \\ 44 & 52 & 11 & 19 & 27 & 35 & 36 \\ 18 & 26 & 34 & 42 & 43 & 51 & 10 \\ 41 & 49 & 50 & 9 & 17 & 25 & 33 \end{bmatrix}$ The sum of all the rows, columns, and diagonals is 224. If anyone is interested, I can explain how to make a $p \cdot p$ magic square for any prime $p \geq 7$.
 March 31st, 2012, 12:42 AM #2 Joined: Mar 2012 Posts: 5 Thanks: 0 Re: 7x7 magic square It kind of reminds me of soduko. How did you do it?
 March 31st, 2012, 09:25 AM #3 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,007 Thanks: 254 Re: 7x7 magic square Here is another . . . [color=beige]. . [/color]$\begin{array}{ccccccc} 38 & 23 & 8 & 49 & 34 & 19 & 4 \\ \\ \\ 30 & 15 & 7 & 41 & 26 & 11 & 45 \\ \\ \\ 22 & 14 & 48 & 33 & 18 & 3 & 37 \\ \\ \\ 21 & 6 & 40 & 25 & 10 & 44 & 29 \\ \\ \\ 13 & 47 & 32 & 17 & 2 & 36 & 28 \\ \\ \\ 5 & 39 & 24 & 9 & 43 & 35 & 20 \\ \\ \\ 46 & 31 & 16 & 1 & 42 & 27 & 12 \end{array}{$ The magic sum is 175.
April 1st, 2012, 09:35 AM   #4
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Joined: Feb 2012

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Re: 7x7 magic square

Quote:
 Originally Posted by JohnLock It kind of reminds me of soduko. How did you do it?
Make an n by n Latin square A as follows:

Start with the top row, which will be all of the numbers 1 to n in order. Now the row below will be the same numbers shifted two places to the right (you can shift any number which is relatively prime to n), and each successive row you will apply the same shift to the previous row. Here is the example for n = 5:

$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 5 & 1 & 2 & 3 \\ 2 & 3 & 4 & 5 & 1 \\ 5 & 1 & 2 & 3 & 4 \\ 3 & 4 & 5 & 1 & 2 \end{bmatrix}$

Now make another Latin square B in the same way, except shifting by a different amount (3 instead of 2, for example):

$\begin{bmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 4 & 5 & 1 & 2 \\ 5 & 1 & 2 & 3 & 4 \\ 2 & 3 & 4 & 5 & 1 \\ 4 & 5 & 1 & 2 & 3 \end{bmatrix}$

Finally, to create the magic square C, each cell $c_{ij}$ will contain the number $n \cdot a_{ij} + b_{ij}$, where $k_{ij}$ represents the entry in the ith row and the jth column of square K.

This will work as long as both shifts are relatively prime to n (in the example 2 and 3 are both relatively prime to 5), and the difference between the shift numbers is relatively prime to n (in the example the difference is 1, which is relatively prime to 5).

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# solution magic 7x7 sum 175

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