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The Magic Cafe Forum Index » » Magical equations » » 5 x 5 Magic Squares (0 Likes) Printer Friendly Version

MagicMan1957
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Hello,
I,m looking for 3 or 4 completed 5 x 5 magic squares for an effect I do.

Not the kind of square where all the rows add up the same but the kind where they choose a number and you cross out all numbers in that row horizontally and vertically leaving only 5 numbers , which add up to the force number.

I can do this myself BUT all 25 numbers must be different, which I cannot figure out how to do. ( when I make the square myself I always end up with duplicate numbers )

And the totals must be under 99, as an example , totals of 88, 67, 73, etc.

Please do not recomend a book to buy, I just need the 3 or 4 squares so I can repeat the effect with different totals for one little effect.

Thank You All!
Scott Cram
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For the basics of constructing a 5x5 forcing matrix, check out Doug Dyment's How to Construct a Forcing Matrix article. You're probably familiar with the steps already, but if you read the article, we'll both be using the same vocabulary.

It's actually fairly simple to make sure that you get 25 different numbers. All you have to do is make sure that the vertical seeds differ by at least 5 (in a 5x5 forcing matrix. It would be 6 in a 6x6 matrix, and so on).

Let me show you what I mean (I'll use periods to keep the proper spacing). First, we'll use 0 through 4 as the 5 horizontal seeds, and step the vertical seeds by 5:

.......0...1...2...3...4
...0
...5
..10
..15
..20

The result when filled in?

........0...1...2...3...4
...0...0...1...2...3...4
...5...5...6...7...8...9
..10.10.1..12.13..14
..15.15.16.17.18.19
..20.20.21.22.23.24

This example is simplified, of course. You don't have to step exactly 5 each row, as long as you step at least 5. Also, the vertical seeds don't have to be in any sort of order. Instead of 0, 5, 10, 15, 20, as in the example, they might read 24, 11, 32, 6, 17.

With the horizontal seeds incrementing by 1, and the vertical seeds increasing by at least 5 each time, you can't get the same number twice. If the horizontal seeds increased by 2, the horizontal seeds would need to increase by 10 each time to prevent duplicate numbers.

A more generalized version of this statement would be that, in an X by X forcing matrix, where the horizontal seeds each differ by a regular increment of Y, the vertical seeds would need to increase by at least X * Y to result in X^2 (X squared) different numbers.

Let's try a new forcing matrix, with the seeds more jumbled:

.......0...4...2...1...3
..24
..11
..32
....6
..17

The result?

...24...28...26...25...27
...11...15...13...12...14
...32...36...34...33...35
.....6...10....8.....7....9
...17...21...19...18...20
MagicMan1957
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Hi Scott,

Yes I've read Doug's article, but you have just simplified the whole thing for me.

Thank you very much for the input!
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