Magic Squares with Magic "Subsquares"

• Jun 4th 2010, 04:04 AM
Samson
Magic Squares with Magic "Subsquares"
Hello All,

I was reading in my text about how one can have a magic square within a magic square. It says that we can form a magic square M that appears in such form:

M = $\begin{pmatrix}
a & a & a & b & b & b & c & c & c\\
a & a & a & b & b & b & c & c & c\\
a & a & a & b & b & b & c & c & c\\
d & d & d & e & e & e & f & f & f\\
d & d & d & e & e & e & f & f & f\\
d & d & d & e & e & e & f & f & f\\
g & g & g & h & h & h & i & i & i\\
g & g & g & h & h & h & i & i & i\\
g & g & g & h & h & h & i & i & i
\end{pmatrix}$

where M itself is a magic square, but M can be broken into ninths and each ninth is also a magic square.

As an example, $\begin{pmatrix}
a & a & a\\
a & a & a\\
a & a & a
\end{pmatrix}$
is a magic square as well which exists inside of M.

Can anyone think of an example where this is the case (in 9x9) ?
• Jun 4th 2010, 09:29 AM
wonderboy1953
Starting off with:

8 1 6
3 5 7
4 9 2

where each row, column and diagonal sums to 15. Now replace the number 1 by the magic square we just started with, replace the number 2 by 2 times the 3 x 3 magic square we just started with, replace the number 3 by 3 times the 3 x 3 magic square we just started with all the way up to replacing the number 9 by 9 times the 3 x 3 magic square we started with. This completes the solution of your problem (I'm curious - which math class are you taking that gave you this problem?)

To mention there's another type of magic square (nested magic square) where peeling off each border leaves you with another magic square.
• Jun 4th 2010, 09:57 AM
Samson
Quote:

Originally Posted by wonderboy1953
Starting off with:

8 1 6
3 5 7
4 9 2

where each row, column and diagonal sums to 15. Now replace the number 1 by the magic square we just started with, replace the number 2 by 2 times the 3 x 3 magic square we just started with, replace the number 3 by 3 times the 3 x 3 magic square we just started with all the way up to replacing the number 9 by 9 times the 3 x 3 magic square we started with. This completes the solution of your problem (I'm curious - which math class are you taking that gave you this problem?)

To mention there's another type of magic square (nested magic square) where peeling off each border leaves you with another magic square.

So here is what I've got:

M = $\begin{pmatrix}
8 & 1 & 6 & 16 & 2 & 12 & 24 & 3 & 18\\
3 & 5 & 7 & 6 & 10 & 14 & 9 & 15 & 21\\
4 & 9 & 2 & 8 & 18 & 4 & 12 & 27 & 6\\
32 & 4 & 24 & 40 & 5 & 30 & 48 & 6 & 36\\
12 & 20 & 28 & 15 & 25 & 35 & 18 & 30 & 42\\
16 & 36 & 8 & 20 & 45 & 10 & 24 & 54 & 12\\
56 & 7 & 41 & 64 & 8 & 48 & 72 & 9 & 54\\
21 & 35 & 49 & 24 & 40 & 56 & 27 & 45 & 63\\
28 & 63 & 14 & 32 & 72 & 16 & 36 & 81 & 18

\end{pmatrix}$

And dear Lord did that take forever to type out lol! I did some preliminary checks, it looks like the squares match up to their own magic squares, but I ran a random check on the diagonal and I get 225, and I get the same thing in a few of the columns, but for column 7 I get 270. Did I screw up somewhere?
• Jun 4th 2010, 10:04 AM
wonderboy1953
Reread the instructions I gave you and ask yourself why should it work.

If you still can't get it, then I or someone else will help you further.
• Jun 4th 2010, 10:09 AM
Samson
Quote:

Originally Posted by wonderboy1953
Reread the instructions I gave you and ask yourself why should it work.

If you still can't get it, then I or someone else will help you further.

I did reread them and I'm sure I'm doing a lot of things right here, I get 225 for the big square for every test I've ran except for col 7 ! What did I do wrong there?
• Jun 4th 2010, 10:11 AM
undefined
You didn't do as wonderboy1953 instructed.

This is the first row:

$\begin{pmatrix}

64 & 8 & 48 & 8 & 1 & 6 & 48 & 6 & 36 \end{pmatrix}$
• Jun 4th 2010, 10:15 AM
wonderboy1953
First two steps
The top 3 x 3 square is:

8 1 6
3 5 7
4 9 2

while the bottom right-hand 3 x 3 square is

16 2 12
6 10 14
8 18 4

(which is the starting 3 x 3 magic square multiplied by 2). Do you see a pattern and can you complete the 7 other subsquares?
• Jun 4th 2010, 10:29 AM
Samson
Quote:

Originally Posted by wonderboy1953
The top 3 x 3 square is:

8 1 6
3 5 7
4 9 2

while the bottom right-hand 3 x 3 square is

16 2 12
6 10 14
8 18 4

(which is the starting 3 x 3 magic square multiplied by 2). Do you see a pattern and can you complete the 7 other subsquares?

I must be screwing it up still, because I can't get consistent values. I get 225, 222, 218....

64 8 48 8 1 6 48 6 36
24 40 56 3 5 7 18 30 42
32 72 16 4 9 2 24 54 16
24 3 18 40 5 30 54 7 42
9 15 21 15 25 35 21 35 49
12 27 6 20 45 10 27 56 14
32 4 24 72 9 54 16 2 12
12 20 28 27 45 63 6 10 14
16 36 8 36 81 18 8 18 4

Sorry its not a nicely formatted :-(
• Jun 4th 2010, 10:32 AM
undefined
Quote:

Originally Posted by Samson
I must be screwing it up still, because I can't get consistent values. I get 225, 222, 218....

64 8 48 8 1 6 48 6 36
24 40 56 3 5 7 18 30 42
32 72 16 4 9 2 24 54 16
24 3 18 40 5 30 54 7 42
9 15 21 15 25 35 21 35 49
12 27 6 20 45 10 27 56 14
32 4 24 72 9 54 16 2 12
12 20 28 27 45 63 6 10 14
16 36 8 36 81 18 8 18 4

Sorry its not a nicely formatted :-(

That should be a 12. There may be other errors too.
• Jun 4th 2010, 10:41 AM
Samson
64 8 48 8 1 6 48 6 36
24 40 56 3 5 7 18 30 42
32 72 16 4 9 2 24 54 12
24 3 18 40 5 30 56 7 42
9 15 21 15 25 35 21 35 49
12 27 6 20 45 10 28 63 14
32 4 24 72 9 54 16 2 12
12 20 28 27 45 63 6 10 14
16 36 8 36 81 18 8 18 4

Voila, I was being an idiot until it dawned on me to use Excel. I was done in like 30 seconds then lol. I totally understand where I went wrong.
• Jun 5th 2010, 05:39 AM
wonderboy1953
Comment
To my knowledge, the largest magic square is 3,001 rows by 3,001 columns (by computer). With the method I just used, I would have expected a far larger magic square by computer.

Anyone want to comment on this point?
• Jun 5th 2010, 06:40 AM
wonderboy1953
BTW Samson
I wasn't trying to be that hard on you. I was following wise advice a college lab professor gave me who, in responding to my question, said, "What do you think?"
• Jun 6th 2010, 09:51 AM
Samson
Quote:

Originally Posted by wonderboy1953
I wasn't trying to be that hard on you. I was following wise advice a college lab professor gave me who, in responding to my question, said, "What do you think?"

It's okay. I'm working my tail off to understand this stuff, so I just hope people know that and I sincerely appreciate the help.
• Jul 8th 2010, 03:49 PM
elemental
I know this thread is long gone, but if the question (and your answer) places no stipulations on repeating integers, why not just create a 9x9 matrix with the same integer as each term?
• Jul 9th 2010, 10:21 AM
wonderboy1953
Quote:

Originally Posted by elemental
I know this thread is long gone, but if the question (and your answer) places no stipulations on repeating integers, why not just create a 9x9 matrix with the same integer as each term?

Trivial (btw this thread will be around for a long time).