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A 2007 Problem (Posted on 2007-01-04) Difficulty: 3 of 5
A magic square of order n is a square array of n consecutive integer numbers (usually, but not neccessarily, from 1 to n) arranged so the sum of the numbers in any horizontal, vertical, or main diagonal line is always the same number, called the magic constant.

For which values of n are there magic squares of order n, with a magic constant of 2007?

See The Solution Submitted by K Sengupta    
Rating: 4.0000 (1 votes)

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Some Thoughts solution comment Comment 2 of 2 |

I was working on a general solution for magic squares with sum of m for positive integers m when I noticed that it is purely coincidental that 3,6, and 9 all provide solutions for the case of m=2007.

if we want magic squares with constant sum of m then we have

(k+t) (t=1 to n^2) = mn

kn^2+(n^2(n^2+1)/2)=mn

2kn^2+n^2(n^2+1)=2mn

obiviously n is not zero so we can divide it out

2kn+n(n^2+1)=2m

n(2k+n^2+1)=2m

so clearly n is a divisor of 2m

now let 2m=r*n for some positive integer r then we have

n(2k+n^2+1)=rn

2k+n^2+1=r

2k=r-n^2-1

k=(r-n^2-1)/2

so for k to be an integer we need that r-n^2-1 be even

now since rn=2m then if n is odd then r must be even otherwise r-n^2-1 would be odd

on the other hand if n is even then r-n^2-1 is even only if r is even as well so we can only have solutions for n when n is a divisor of 2m, less than the cube root of 2m, and if n is odd then 2m/n must be  even.

So for example if m=100 then we have 2m=200, cube root of 2m is 5.84 so n must be less than or equal to 5 and a divisor of 2m, that leaves 1,2,4, and 5.  obviously 1 is a solution and 2 is not as shown by K Sengupta.  But 4 is not a solution because 200/4=50 and we would have 4(2k+17)=200 or 2k+17=50 or 2k=33 or k=16.5 and thus does not give an integer solution for k.  So for m=100 there are magic squares that sum to 100 only for n=1 and n=5.


  Posted by Daniel on 2009-02-02 21:36:51
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