Regarding the "magic square"?

Question

In the book, "the man who counted", there's a magic square made up by 9 small squares. The point of this square is, by using the number 1-9 only once, to have the sum of each line or column or diagonal always add up to 15.

Now, that may not be all that difficult, but my friend says there is a special pattern in which you enter the numbers in these squares. I wanted to find out myself, so... my question is what kind of math problems/exercises would help someone think "more mathematically", if you will, and prepare me to solve something like this?... any other info on this subject is welcome..

Answer 1

Your friend may be saying that there is essentially only one way to create a 3x3 magic square, with all other solutions being rotations or mirror images of the square. If so, he's right about that. Proving it doesn't require thinking "more mathematically" - it's mostly just deductive logic. I worked out a proof once... ah, here it is...


Discounting rotations and mirror images, there is only one way to make a magic square using the digits 1 through 9.

First, we can prove the 5 has to go in the middle:

Let the numbers in the top row from left to right be a, b, and c. Let the numbers in the middle row from left to right be d, e, and f. (Note e is the middle box.) Let the numbers in the bottom row from left to right be g, h, and i.

We know a + b + c = 15 and g + h + i = 15

We also know a + e + i = 15, b + e + h = 15, and c + e + g =15. Therefore, (a + e + i) + (b + e + h) + (c + e + g) = 15 * 3 = 45. We can rearrange this to (a + b + c) + (g + h + i) + 3e = 45. Since a + b + c = 15 and g + h + i = 15, that reduces to 15 + 15 + 3e = 45, or 3e = 15. Therefore e (the middle box) must equal 5.

Knowing the middle box is 5, try 9 in a corner and you'll notice a problem - the 6, 7 and 8 can't all be placed in the square without the row or column with the 9 exceeding 15. So, the 9 has to occupy a side box. Since we are discounting rotations and mirror images, we can arbitrarily select box b for the 9. This forces the placement of the 1 in box h.

With the 9 and 1 now placed, the 8 and 6 have to go in the row with the 1. So, we'll put the 8 in box g and the 6 in box i. (We still haven't done anything to create a second square if rotations and mirror images are discounted.) That forces the placement of the final four numbers.

Answer 2

look, acctually I had a similar question few years ago regarding this square and this pattern is not only applicable to 3x3 square but larger squares as well

Suppose that you have the numbers from 1-9, the only thing you have to do is the following: take a paper and write the numbers in diagonal and leave some spaces (I use "-" signs just to arrange the numbers)

------3
---2----6
1----5----9
---4----8
------7

If you look, you have in the center a square with corners 2, 6, 4, 8

2----6
---5--
4----8

this is your magic square. how to arrange the other numbers? Just write them in the same vertical/horizontal line they appear but in the space at the maximum distance, for example, the number 1 between 6 & 8 and so on to give you

2 7 6
9 5 1
4 3 8

It's now easy!! Now, what happen if you want to do it with numbers 2-10 or 5-13, you have to do the same and the square gives you the same properties. Instead of begin to write the number 1, write the number you want! and in diagonal in the same form as above write the next numbers and follow the instructions. More...

What if I don't want to have numbers separated by 1? Can I put the numbers 2, 4, 6, and so on? Yes, try it in the same way. At this point we are able to make a 3x3 magic square with any 9 numbers separated by a equal distance, no matter what distance is. (you may construct a square in which the numbers are separated pi for example)

there's more, give me a little more time, now I have to take an hourto do something, but I will write more of this answer if you have a little more time, it's interesting

Answer 3

Hard to know for certain what your friend was speaking of, but here is another possibility. For any odd number n, you can make an nxn magic square in the following way (using the numbers 1 ... n^2, with common sum n*(n^2 + 1)/2).

Start with 1 in the bottom middle square, then moving down and to the right place the rest of the numbers in order. When you move off the bottom of the magic square go back to the top, and when you move off the right of the square go to the far left, and when you run into another number go up a space (go to the bottom if you are already at the top).

Here's how it works for a 3x3:
_ _ _
_ _ _
_ 1 _

_ _ 2
_ _ _
_ 1 _

_ _ 2
3 _ _
_ 1 _

4 _ 2
3 _ _
_ 1 _

4 _ 2
3 5 _
_ 1 _

4 _ 2
3 5 _
_ 1 6

4 _ 2
3 5 7
_ 1 6

4 _ 2
3 5 7
_ 1 6

4 _ 2
3 5 7
8 1 6

4 9 2
3 5 7
8 1 6

And we are done!!!

Now try it with a 5x5

Answer 4

Sounds like Soduku!
Recently I obtained an ebook, written by a retired math teacher, explaining how to solve Sodukus systematically, using the pigeonhole principle. He never explicitly mentions the pigeonhole principle but I immediately recognised it when I read his book and applied his technique. We're not supposed to promote commercial sites here but if you want a link to this ebook, email me.
As for how to lay out the numbers to create a magic square or sudoku from scratch, there is a pattern for that but I havn't figured it out yet. Let's see what the other guy says...

Answer 5

i would not have confidence the words of Anton Lavey approximately something. If I keep in mind wisely he became a follower of the daddy of Lies. besides the shown fact that to that end Anton does believe the Apostle Paul asserting that one and all such deceptions are from devil. i'm undecided there is taken into consideration one of those element as "White Magic" or "Black". In any experience the Bible says stay faraway from all of it. Pastor artwork