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# Magic fractions

Fri 10 Dec, 2004 03:49 am
5/36, 7/36, 1/18, 5/18, 1/12, 1/9, 2/9, 1/6, 1/4

Put the above fractions into a 3x3 magic square
such that every line adds up to 1/2.

Whim
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Type: Discussion • Score: 1 • Views: 2,693 • Replies: 7
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markr

1
Fri 10 Dec, 2004 10:09 am
1/12 2/9 7/36
5/18 1/6 1/18
5/36 1/9 1/4
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Perdition

1
Sat 11 Dec, 2004 09:24 am
how is it you plan a magic square without trying random combonations btw?
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markr

1
Sat 11 Dec, 2004 12:33 pm
Here's one way for odd sized squares. I'll use a 5x5 as an example.
Starting with 1, write the numbers on diagonals as shown. The x's are gaps to be filled later. Draw a box around the 5x5 middle part. Slide the groups of numbers outside each side of the box across to the opposite side (inside) to fill in the gaps. The numbers that slide stay in the same positions relative to each other. I did the top and right sides in the example. Do the same for the bottom and left.
Code:``` 05 04 xx 10 ---------------- |03 xx 09 xx 15| 02|xx 08 xx 14 xx|20 01 xx|07 xx 13 xx 19|xx 25 06|xx 12 xx 18 xx|24 |11 xx 17 xx 23| ---------------- 16 xx 22 21 xx (top outside numbers moved to bottom inside) ---------------- |03 xx 09 xx 15| 02|xx 08 xx 14 xx|20 01 xx|07 xx 13 xx 19|xx 25 06|xx 12 05 18 xx|24 |11 04 17 10 23| ---------------- 16 xx 22 21 xx ---------------- |03 xx 09 xx 15| 02|20 08 xx 14 xx| 01 xx|07 25 13 xx 19|xx (right outside numbers moved to left inside) 06|24 12 05 18 xx| |11 04 17 10 23| ---------------- 16 xx 22 21 ```
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Perdition

1
Sun 12 Dec, 2004 03:55 am
Not half bad question is how would you apply this if you wanted to make a certain magic sum such as 38? Making the values along the diagnals sequentially will only supply one possible square not to mention one specific magic sum
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markr

1
Sun 12 Dec, 2004 10:37 am
During construction, you don't have to start with 1 in the first position. Also, after you're done, you can add a constant to all the numbers. You may not be able to get what you want by limiting yourself to integers.
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Perdition

1
Sun 12 Dec, 2004 03:29 pm
well after making a couple of squares of order 3, 5 ands 7, i see a pattern as to how one can choose the magic sum to one's liking for a specific odd order.
Let us say that the first number you start off with, as mentioned before, will be symbolised with b (base). let us state the order as x.

magic sum= b*x+x( x^2 -1 /2) = x(b+(x^2-1)/2)

so if you wanted to start the process that mark mentioned with the first number being 3, for an order of 5, the magic sum once it is done would be:
5(3+(5^2-1)/2) = 5(3+12)=75, whereas if the base number was 1 the magic sum would be 5(1+12)=65. kind of cute I suppose, reminds me of the equation of an arithmetic series although i don't recall it precisely so that may be inaccurate. if anyone has anything else to add to this fun fun investigation into the art form which are magic squares, they're great for parties, jump right in with a method for even orders hehe
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markr

1
Sun 12 Dec, 2004 08:02 pm
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