The past Monday, (April 5, 2004) Rodolfo Kurchan sent to me by email the following nice 5x5 antimagic square:

The stunning feature of this antimagic 5x5 square is that contains an embedded centered nut of a 3x3 magic square (numbers in red color); that is to say here we have a magic square inside an antimagic square! A beauty example of an object that contains within its contrary.

Perhaps the beauty of kind of this objects is excuse enough in order to keep away the prime numbers for this puzzle.

Questions:

1. Can you get another antimagic solution of the same size (5x5) using a distinct magic (3x3) square?

2. Can you get an antimagic 5x5 containing an eccentric (non-centered) magic 3x3

3. Can you get a larger example (i.e. an antimagic 6x6 containing a magic 4x4)?

4. Can you get the opposite concept: a magic square containing inside an antimagic square (*)

_______

(*) It has been shown that no antimagic square of order less than 4 can exist; then, the minimal example of this kind of objects could be an magic square 6x6 containing a centered antimagic square 4x4.

Solution:

I feel my self really happy because the readers of my pages are more daring than I suppose. While I was thinking that this puzzle was hard enough in order to add the primality condition of the numbers used, J.C. Rosa got a solutions to Question 4 using only primes! Here is what he wrote:

It is possible to find one 3x3 antimagic square with prime

numbers ( see Won plate 132 ) and particularly

for your puzzle 263 I have found this :

 

        The following antimagic square is composed of nine primes
 with its eight sums in arithmetic progression (step 2 ). The sums
 go from 443 up to 457:
 
                       101    113    233
                       293    151     13
                        59    191    199
 
(note that the central number is a palprime )
 
And now the same antimagic square embedded in a 5x5 magic square:
 
                29     157    277    263      23
               229     101    113    233      73
                11     293    151     13     281
               197      59    191    199     103
               283     139     17     41     269
 
                magic sum = 749
 
     (note that this puzzle is the puzzle 263 and 263 is inside this
square)

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J. C. Rosa added:

do you want an 3x3 prime antimagic square

embedded in a 5x5 prime magic square with a prime magic sum ?

 

Here it is ::
 
                83      43     139       23      101
              151     29    113       89         7
                41     149     79       13     107
                53      59      47      127     103
               61     109     11     137       71

 

For the 3x3 antimagic square the sums go from 227 up to 241 ( step 2). For the 5x5 magic square the magic sum is 389 (prime )

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J. C. Rosa also contributes to Q1:

About the question 1 of the puzzle 263 there are a lot of different

solutions . Here are two examples with the numbers from 1 to 25

and one example with only 25 prime numbers ( unfortunately they are

not consecutive ).See below. Now I hope to find a bridge between the 

question 1 of the puzzle 263 and the question 2 of the puzzle 264 :

A 3x3 magic square embedded in an 5x5 antimagic square composed

only 25 consecutive primes....

 

  5     7     20     18     10

17     2     23      14     6

 4     25    13      1      21

19    12     3       24    11

16    22     8       9      15

 

Magic sum=39 . The sums go from 59 to 70

 

              xxxxxxxxx

 

  5    14     18     17     8

15     2     21      13     10

 4     23    12      1      25

24    11     3       22     7

20    16     9       6      19

 

Magic sum=36 . The sums go from 59 to 70

 

               xxxxxxxxxxxxx

 

67     31       23      107      61

 7      17       89       71      109

41     113      59         5        83

127    47       29       101      3

53      97       103      13       43

 

Magic sum=177    The sums go from 287 to 309 (step 2)

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J. C. Rosa contribution to Q2 arrived the 12/6/04:

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Anurag Sahay had previously sent (Jan 2005) the following solutions to Q3:

> Some solutions for Q3:
>
> 5 32 29 3 6 30
> 27 12 19 18 25 8
> 36 26 24 11 13 7
> 9 15 17 22 20 31
> 34 21 14 23 16 2
> 4 10 1 35 33 28
>
> 7 29 34 3 9 28
> 30 12 19 18 25 10
> 6 26 24 11 13 31
> 35 15 17 22 20 8
> 33 21 14 23 16 2
> 5 1 4 36 32 27
>
> 6 7 34 29 32 2
> 33 12 19 18 25 10
> 4 26 24 11 13 36
> 8 15 17 22 20 27
> 30 21 14 23 16 1
> 31 35 5 3 9 28
>
> 4 3 34 36 31 2
> 28 11 25 12 26 9
> 1 22 20 17 15 30
> 8 23 13 24 14 35
> 33 18 16 21 1
 

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