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Perfect magic squares
pan magic 9x9 square

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How to produce a 9x9 pan magic square
 
 
I don’t know how many pure (pan)magic 9x9 squares exist.
 

A 9x9 square is an odd square, but is also a multiple of 3 (= can be devided by 3).
Can a 9x9 panmagic square be produced by using the same method of construction as the method to produce a
5x5 panmagic square (which can be used to produce all possible panmagic 5x5 and panmagic 7x7 squares)?
The answer is yes and no. If you choose as first row the digits 0-1-2-3-4-5-6-7-8 you get as result only a semi-
magic 9x9 square. If you choose as first row the digits 0-2-1-5-4-3-7-6-8 you get as result a correct pan-
magic 9x9 square.

Notify first that the row 0-2-1-5-4-3-7-6-8 leads to a correct result, because 0+5+7, 2+4+6 and 1+3+8 is 12,
that is 1/3 of (0+1+2+3+4+5+6+7+8=) 36.

Notify second that if you choose as first row the digits 0-2-1-5-4-3-7-6-8 you can, instead of the 2nd square that
is produced by shifting the first row to the right, take as 2nd square the 1st square rotated by a quarter to the right!
 
 
There is a method to produce a panmagic 9x9 square plus an extra magic feature:
Fill in the first row of the 1st square: 0, 1, 2, 3, 4, 5, 6, 7 and 8. Produce the second and third row of the 1st square
by moving the digits each time 3 places to the left.

 
                                             1st square, first three rows
 
 
 
 
 
 
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
8
 
 
 
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
8
 
 
 
0
1
2
3
4
5
6
7
8
0
1
2
3
4
5
6
7
8
 
 
 
 
 
 
 
 
The first three rows of the 1st square consist of three 3x3 sub-squares. Produce the second three rows of the 1st
square by switching the sequence of the three columns of the three 3x3 sub-squares to 2-3-1. Produce the second
three rows of the 1st square by switching the sequence of the three columns of the three 3x3 sub-squares to 3-1-2.
 
 
 1st square
0
1
2
3
4
5
6
7
8
3
4
5
6
7
8
0
1
2
6
7
8
0
1
2
3
4
5
1
2
0
4
5
3
7
8
6
4
5
3
7
8
6
1
2
0
7
8
6
1
2
0
4
5
3
2
0
1
5
3
4
8
6
7
5
3
4
8
6
7
2
0
1
8
6
7
2
0
1
5
3
4
 
 
Produce the 2nd square by rotating the 1st square (a quarter turn to the right). Take a digit from the 1st square multi-
plied by 9 and add (1x) the digit from the same cell of the 2nd square.
 
 
  9x digit                                              +   1x digit                                               =    panmagic 9x9 square
0
1
2
3
4
5
6
7
8
 
8
5
2
7
4
1
6
3
0
 
8
14
20
34
40
46
60
66
72
3
4
5
6
7
8
0
1
2
 
6
3
0
8
5
2
7
4
1
 
33
39
45
62
68
74
7
13
19
6
7
8
0
1
2
3
4
5
 
7
4
1
6
3
0
8
5
2
 
61
67
73
6
12
18
35
41
47
1
2
0
4
5
3
7
8
6
 
2
8
5
1
7
4
0
6
3
 
11
26
5
37
52
31
63
78
57
4
5
3
7
8
6
1
2
0
 
0
6
3
2
8
5
1
7
4
 
36
51
30
65
80
59
10
25
4
7
8
6
1
2
0
4
5
3
 
1
7
4
0
6
3
2
8
5
 
64
79
58
9
24
3
38
53
32
2
0
1
5
3
4
8
6
7
 
5
2
8
4
1
7
3
0
6
 
23
2
17
49
28
43
75
54
69
5
3
4
8
6
7
2
0
1
 
3
0
6
5
2
8
4
1
7
 
48
27
42
77
56
71
22
1
16
8
6
7
2
0
1
5
3
4
 
4
1
7
3
0
6
5
2
8
 
76
55
70
21
0
15
50
29
44
 
 
In the square consists of the digits 0 up to 80. By adding 1 to each digit you can produce a square which consists of the
digits 1 up to 81.


Extra magic feature is that the sum of the digits of each 3x3 sub-square is the magic sum of 360.
 
 
You can use this method to produce squares which have the size of odd number squared (= 9x9, 25x25, 49x49, 81x81,
121x121, ...). For example, to produce a 25x25 square use the digits 0 up to 24 as first row of the 1st square. Produce
the second up to fifth row of the 1st square by moving the digits each time 5 places to the left. Produce the second up to
fifth five rows (= 5x5 sub-squares) of the 1st square by switching the sequence of the five columns of the five 5x5 sub-
squares to 2-3-4-5-1, 3-4-5-1-2, 4-5-1-2-3 respectively 5-1-2-3-4.


There are not many possibilities (just a little alternative combinations of digits) to produce a 9x9 panmagic square. You
can not find much information about 9x9 panmagic squares on other websites. Try www.grogono.com/magic/9x9.php


See also:  Panmagic 15x15 square



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