Largest Magic Square Ever KnownDate: 09/18/2001 at 03:46:49 From: Jayson Javellana Subject: Largest Magic Square Ever Known Dr. Math, I am eager to know what was the largest magic square ever constructed and known in the world of mathematics. What is the easiest way to find the sum of each row, column, and main diagonals of magic squares? Thank you. Jayson Date: 09/18/2001 at 11:35:02 From: Doctor Rob Subject: Re: Largest Magic Square Ever Known Thanks for writing to Ask Dr. Math, Jayson. There are magic squares of every odd order, so there is no maximum. Here is a 9-by-9 magic square that shows the pattern for the construction of odd orders (see where the numbers 1, 2, ..., 9 are placed; then 10, 11, ..., 18; then 19, 20, ..., 27; and so on): 37 28 19 10 1 73 64 55 46 47 38 29 20 11 2 74 65 56 57 48 39 30 21 12 3 75 66 67 58 49 40 31 22 13 4 86 77 68 59 50 41 32 23 14 5 6 78 69 60 51 42 33 24 15 16 7 79 70 61 52 43 34 25 26 17 8 80 71 62 53 44 35 36 27 18 9 81 72 63 54 45 There are also known constructions for large even orders, but they are more complicated. If an n-by-n magic square contains the numbers from 1 to n^2, then the sum of any row, column, or main diagonal must be n*(n^2+1)/2. For example, if n = 9, as above, then the sum would be 9*82/2 = 369. - Doctor Rob, The Math Forum http://mathforum.org/dr.math/ Date: 09/18/2001 at 11:48:07 From: Doctor Jubal Subject: Re: Largest Magic Square Ever Known Hi Jayson, Thanks for writing to Dr. Math. I searched the Dr. Math archives on the keyword: magic square (exact phrase) and found this link Finding Magic Squares http://mathforum.org/dr.math/problems/holaday5.12.98.html which gives instructions for creating a magic square as large as you want, so long as the square has an odd number of squares on a side. Actually, however, you can create magic squares as large as you want, even if they have an even number of squares on a side. If the number of squares on each side is divisible by four, start off by just numbering the squares in order across each row: (I'll illustrate the method for 8x8.) 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 Now, divide the square into smaller 4x4 squares, and draw in the diagonals of those four-by-four squares. Some of the numbers will be crossed out in this process \ 2 3 / \ 6 7 / 9 \ / 12 14 \ / 16 17 / \ 20 21 / \ 24 / 26 27 \ / 30 31 \ \ 34 35 / \ 38 39 / 41 \ / 44 45 \ / 48 49 / \ 52 53 / \ 56 / 58 59 \ / 62 63 \ Finally, replace all the numbers that got crossed out with n^2 + 1 - a, where n^2 is the number of cells in the square (64 in this case) and a is what the number originally was. This gives you 64 2 3 61 60 6 7 57 9 55 54 12 14 51 50 16 17 47 46 20 21 43 42 24 40 26 27 37 36 30 31 33 32 34 35 29 28 38 39 25 41 23 22 44 45 19 18 48 49 15 14 52 53 11 10 56 8 58 59 5 4 62 63 1 which is a magic square. This will hold for any square of side n divisible by 4. There is also a way to generate magic squares as large as you want with sides of 4n+2 (even, but not divisible by 4), but it's more complicated. Does this help? Write back if you'd like to talk about this some more, or if you have any other questions. - Doctor Jubal, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/