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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?

  Submitted by K Sengupta    
Rating: 4.0000 (1 votes)
Solution: (Hide)
In the first line, “n2 consecutive integer numbers” should read as: “n2 consecutive positive integer numbers”. The problem text was changed at the queue, and it was an omission on my part not to add the word "positive" in the said revision.

SOLUTION:

At the outset, if any 2x2 magic square were feasible then let a and b denote the elements of the first row and a and c denote the elements of the first column. Then, by the properties of magic square we obtain: a+b = a+c; giving: b=c, which is a contradiction.

However, we know that the simplest magic square is the 1x1 magic square whose only entry is the number 1. In the current problem, the only entry for the order 1 magic square is the number 2007.

Accordingly, the nxn magic square in our problem is such that n is not equal to 2 .....(#)

Let the nxn magic square consist of the elements: k+i, for i = 1,2,...., n^2

Then by the problem:

(k+1) + (k+2) + ……..+ (k+ n^2) = 2007*n
Or, k*n^2 + n^2(n^2+1)/2 = 2007*n
Or, n(2k+ (n^2+1)) = 4014………..(i)
Since n must be a positive integer, it follows that n must divide 4014.

Again, 4014 = n(2k+ (n^2+1)) > n^3
Or, n^3 < 4014 < 4096 = 16^3
Or, n < 16

Clearly the only positive integer divisors of 4014 less than 16 are 1, 2, 3, 6, 9.

However, by (#) no 2x2 magic square is feasible.

Consequently the required values of n for which there exist magic squares of order n with common row-, column-, and main diagonal-sum 2007 are 1, 3, 6 and 9.

NOTE: If k=0, then n^3 + n = 4014, which does not possess any solution in positive integers. Hence, no magic square of order n with the first n^2 natural numbers having a magic constant of 2007 is feasible.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some Thoughtssolution commentDaniel2009-02-02 21:36:51
Some ThoughtsOnly positive numbers. Don't ask me to actually find them.Jer2007-01-05 13:33:38
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