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The basic pattern is the magic square of 0 (as magic sum).
|
22
|
14
|
-46
|
10
|
|
-30
|
-6
|
6
|
30
|
|
26
|
2
|
-2
|
-26
|
|
-18
|
-10
|
42
|
-14
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If you choose as magic sum 403, then 403 : 4 = 100 remainder 3.
Add 100 to each cell of the basic pattern of 0 and also add additionally the remainder number (3) to each
yellow marked cell.
|
125
|
114
|
54
|
110
|
|
70
|
94
|
109
|
130
|
|
126
|
102
|
98
|
77
|
|
82
|
93
|
142
|
86
|
This square is not panmagic, but has other additional magic features (for example the sum of the digits
of the four 2x2 subsquares, the four horizontal twisted rows and the four vertical “ears” = 403).
Check it out for yourself.
Is there an alternate method, which is even better?
There is a mathematical alternative with the following basic key:
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a
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c+3
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d+1
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b+2
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d+2
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b+1
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a+3
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c
|
|
b+3
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d
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c+2
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a+1
|
|
c+1
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a+2
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b
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d+3
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Notify that: a + b + c + d = magic sum - 6
Boundary condition is: b - a >= 4 and c - b >= 4 and d - c >= 4 (to get 16 different digits)
Choose a magic sum, for example 100. Take care that a + b + c + d = 100 - 6 = 94
(and that there is a minimal difference of 4 between a and b, b and c, and c and d), for example:
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a =
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7
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b =
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20
|
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c =
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31
|
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d =
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36
|
|
7
|
34
|
37
|
22
|
|
38
|
21
|
10
|
31
|
|
23
|
36
|
33
|
8
|
|
32
|
9
|
20
|
39
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N.B.: Establish that the square is "Dürer" magic (because it has the same magic features as the Dürer
magic square).
It is also possible to produce a panmagic 5x5 square for each random sum; use for example the follo-
wing table:
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a
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b+1
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c+2
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d+3
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e+4
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|
c+3
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d+4
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e
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a+1
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b+2
|
|
e+1
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a+2
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b+3
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c+4
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d
|
|
b+4
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c
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d+1
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e+2
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a+3
|
|
d+2
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e+3
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a+4
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b
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c+1
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