Friday

October 28, 2016
Posted by **Logan** on Wednesday, September 26, 2007 at 1:10pm.

- 15 - 6 1 -

33 - - - - -

- 28 - - 34 7

24 16 - 27 - 21

14 - 5 26 - -

- 35 18 31 - -

- Math -
**tchrwill**, Wednesday, September 26, 2007 at 1:52pmI was given the following magic square and was to fill in the blanks. I know that all the rows and center diagonals are to add up to 111 and no number is used more than once (1-36). I just can't figure it out. I can get all but 2 columns and one diagonal. Help!

First you must learn how to create a 3x3 cell magic square. Then you will have to follow another procedure to create the 6x6 cell magic square.

Before getting into the mechanical or systematic methods of creating magic squares, let me first show you that it is possible to create one by means of simple logic, although probably only for the 3 cell square.

The usual 3 cell magic square problem is posed as, "Place the numbers 1 through 9 in the 9 cells of the 3x3 square such that each row, each column, and each diagonal add up to the same total."

Of course, the typical trial and error approach will ultimately get you to an answer but the more rewarding method is your own intuition and logic. Lets see where this takes us.

The first thing you might ask yourself is what is the total that we are seeking with the 9 digits. Since all three rows or columns must add up to the same total, it stands to reason that the sum of the rows or columns must, by definition, add up to the sum of the 9 digits, which turns out to be 45. Therefore, each row or column must add up to 45/3 = 15.

You might notice that 8 of the digits we are using just happen to add up to 10, 1+9, 2+8, 3+7, and 4+6. It might also occur to you that the middle number of the 9 digits, the 5, would most logically want to be in the middle cell with the others located around it all adding up to 15. Where to start?

What if the 9 were located on a corner? Since all three lines of numbers, including that corner 9, must toal 15, we would need three pairs of numbers that each add up to 6. Of course, this is impossible as we only have 1+5 and 2+4 at our disposal thus forcing the 9 to be located in the middle cell of one of the sides. Lets try the middle

cell of the bottom row (it could be any of the four available positions) which forces the 1 to be in the middle cell of the top row.

.................................? 1 ?

.................................? 5 ?

.................................? 9 ?

Looking at our bottom row now, we notice that the two outer numbers must add up to 6 and we only have 2+4 available to us. For a reason that will become obvious later, lets try the 2 in the lower right hand corner and the 4 in the lower left hand corner.

.................................? 1 ?

.................................? 5 ?

.................................4 9 2

What do you know? It looks like our intuition can take a rest now as the other seem to all just fall into place. The upper left cell must be an 8, the upper right must be a 6, and of course, the middle left is forced to be a 3 and the middle right becomes the 7. Here we are, and just by thinking it through.

.................................8 1 6

.................................3 5 7

.................................4 9 2

Note that the outer numbers can be rotated clockwise or counterclocwise to define 7 additional arrangements. Mirror imaging about both vertical and horizontal axes as well as the diagonal axes will produce more. See how many different ones you can define overall.

Now we will get back to the more traditional method.

The simplest magic square has 3 cells on a side, or 9 cells altogether. We call this a three square. The simplest three square is one where you place the numbers from 1 to 9, inclusive, in each cell in such a way that the sum of every horizontal and vertical row as well as the two diagonal rows add up to the same number. This basic 3 cell square, adding up to 15, looks like the following (looks familiar):

8 1 6

3 5 7

4 9 2

This 3 cell square has some other strange characteristics. All of the four lines that pass through the center are in arithmetic progressionhaving differences of 1, 2, 3, and 4. Notice also that the squares of the first and third columns are equal, i.e., 8^2 + 3^2 + 4^2 = 6^2 + 7^2 + 2^2 = 89. The sum of the middle column squares is 1^2 + 5^2 + 9^2 = 107 which is equal to 89 + 18. The sum of the squares in the rows total 101, and 83 and, strangely enough, 101 - 83 = 18.

Other three cell magic squares can be created where the rows all add up to other numbers, the only constraint being that the sums of the rows must be divisible by 3. For instance, magics square adding up to 42 and 48 would look like this:

17 10 15 22 8 18

12 14 16 12 16 20

13 18 11 14 24 10

While we have only looked at three cell magic squares so far, you might have noticed a couple of things that turn out to be fundamental to all odd cell magic squares. First, the center square number is the middle number of the group of numbers being used or the sum of the first number and the last number divided by 2; it is also equal to the row total divided by the number of cells in the square.

Your next logical question is bound to be,"How does one create such squares? Well, I will try to explain it in words without a picture.

First draw yourself a three cell square with a dark pencil and place the numbers 1 through 9 in them as shown above. Now, above squares 1 and 6 draw two light lined squares just for reference. Similarly, draw two light lined squares to the right of squares 6 and 7. Their use will become obvious as we go along. Always place the first number being used in your square in the top center square, number one in our illustration. Now comes the tricky

part. We now wish to locate the number 2 in its proper location. Move out of the number 1 box, upward to the right, at a 45 degree angle, into the light lined box. Clearly this imaginary box is outside the boundries of our three cell square. What you do is drop down to the lowest cell in that column and place the 2. Now for the 3, move upward to the right again into the light lined box next to the number 7. Again you are outside the three cell square

so move all the way over to the left in that row and place the 3. Now you will notice that you cannot move upward to the right as you are blocked by the number 1. Merely drop down one row and place the 4 directly below the 3. Move upward to the right and place the 5. Again, move upward to the right and place the 6. You now cannot move upward to the right as there is no imaginary square there for you to move into. Merely drop down one cell and place

the 7 directly below the 6. Now move upward and to the right again and you are outside the square again. As before, merely move all the way over to the left cell in that row and place the 8. Moving upward and to the right again, you are outside the square again so merely drop down to the bottom cell in that column and place the 9. This exact same pattern is followed no matter what the rows and columns add up to.

You can also enter the numbers starting in the center box of the right column, the center box of the bottom row, or the center box of the left column as long as you follow the same pattern of locating numbers. If you were to do this you would end up with the following squares:

4 3 8.........2 9 4..........6 7 2

9 5 1.........7 5 3..........1 5 9

2 7 6.........6 1 8..........8 3 4

Moving all the outer numbers one or more boxes clockwise, or counterclockwise, also produces the same result.

Your next question is bound to be, "How does one create a magic 3 square that adds up to something other than 15?" There are two ways to create magic square for your friends. First, ask them for a number, say no more than 2 digits. You then proceed to place their number in the top center cell and continue to fill in the square, in the same basic pattern we just described above, telling them that when you are done, the rows, columns, and diagonals will all add up to a specific number.

The second way to create a magic square is to ask them for a number larger than 15 that is divisible by three. You then proceed to fill in all the cells in the same pattern such that the rows, columns, and diagonals add up to the number they gave you. Here is how you do it.

In the first method, ask them for a number, say from 1 to 25, but it can be any number. Lets say they give you 17. In your head, multiply 17 times 3 and add 12, such that 3(17) + 12 = 63. You now place the number 17 in the top center cell, continue to place 18, 19, 20, 21, 22, 23, 24, and 25 in the the cells in the same pattern as you placed the numbers 1 through 9 above. As you are doing this, you tell them that when you are done, every row, column, and the two diagonals will add up to 63. What magic.

24 17 22

19 21 23

20 25 18

For the second method, ask them for a number that is divisible by three as you are working with a 3 x 3 square. Lets say they give you 48. In you head now, subtract 12 from the number they give you and divide the result by 3. For our example you will get 12. So place the number 12 in the top center cell and continue to fill in the other

cells in the same pattern until you reach the last cell with the number 20. Lo and behold, every row, column, and the diagonals, add up to 48.

19 12 17

14 16 18

15 20 13

Now, as strange as it might seem, even though a 6 cell magic square is an even number square, it is not created in the manner just described. The 6 cell square is considered the most difficult magic square to create. I give you what has been advertised as the easiest method discovered to date, as far as I know.

First, divide the 6 cell square into four 3 cell squares. Starting with the upper left 3 cell square, place the numbers 1 through 9 in their proper cells exactly as you did in the regular 3 cell square. Now, moving to the lower right 3 cell square, place the numbers 10 through 18 in exactly the same pattern as you placed the 1-9 numbers.

Moving to the upper right 3 cell square, place the numbers 19 through 27 in this 3 cell square in the same pattern. And lastly, in the lower left 3 cell square, place the numbers 28 through 36 in the same pattern. Now the tricky part. Remove the numbers 8, 5, and 4 from the upper left 3 cell square, holding them aside. Move the

numbers 35, 32, and 31 from the lower left 3 cell square to the three empty spaces just created. Replace the numbers 35, 32, and 31 with the numbers 8, 5, and 4 from the cell above. Got it? You now have a 6 cell square that adds up to 111 in all the rows.

Good luck.