Question 987076

Create a 3-by-3 magic square using nine of the ten numbers 20, 21, 22, 23, 24,
25, 26, 27, 28, and 29.  Explain your solution and reasoning.  List the
strategies you have used. 
<pre>

21 28 23
26 24 22
25 20 27

and

22 29 24
27 25 23
26 21 28

Here is my strategy:

In base 3 there are only 3 digits, 0, 1, and 2

I can easily make a magic square using those 3 digits on
each column and row, like this:

0 2 1    
2 1 0
1 0 2 

The rows, columns and diagonals all add up to 3 

And if I rotate it 90° clockwise I have this magic square,
where all of the digits changed position except the middle number.

1 2 0
0 1 2
2 0 1

And if I put the digits of those two magic squares together,
I get this:

01 22 10
20 11 02
12 00 21

I notice that all those are different integers in base 3

Now if I take those and convert them to base 10 numbers, I will
get a magic square because the 3's digits and the 1's digits
add to the same number in each row. column, and diagonal.

0×3+1  2×3+2  1×3+0
2×3+0  1×3+1  0×3+2
1×3+2  0×3+0  2×3+1

which is

1 8 3
6 4 2
5 0 7

As we see, all the rows, columns and diagonals add up to 12. 
So that's a magic square with the integers from 0 through 8.

So I can make that into a magic square with the integers from 
20 through 28 just by adding 20 to each number:

1+20=21 8+20=28 3+20=23
6+20=26 4+20=24 2+20=22
5+20=25 0+20=20 7+20=27

which is this magic square

21 28 23
26 24 22
25 20 27

All the rows, columns and diagonals add up to 72.

-------

Or I could have made that into a magic square with the integers from 21 
through 29 by adding 21 to each number instead of 20. That would have 
just made each number 1 more than they are in the above magic square:

22 29 24
27 25 23
26 21 28

And all the rows, columns and diagonals add up to 75.

Edwin</pre>