Magic sum of a normal magic square
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The magic sum of a normal magic square of order n
In my previous blog post, I described a normal magic square of order 3, i.e. a 3x3 array of numbers. The magic sum of this magic square is 15.
But which is the magic sum of a normal magic square in the general case?
Well, we can use the same calculation principle as for the 3x3 magic square!
For a normal magic square the number series starts with a '1' and consists of natural numbers in an increasing sequence.
The magic square of order n, consists of n rows and n columns, i.e. there are n^{2} numbers.
First we calculate the sum 'S' of all numbers in the magic square. To do this, we can use a little trick that you might remember from school.
In the first row of the table below, we put all numbers in increasing order. The sum of this row is S.
In the second row of the table, we put all numbers in decreasing order. The sum of this row is of course also S, since it is the same numbers but in a different order.
S= | 1 | 2 | 3 | ... | n^{2} |
S= | n^{2} | n^{2}-1 | n^{2}-2 | ... | 1 |
2*S= | n^{2}+1 | n^{2}+1 | n^{2}+1 | ... | n^{2}+1 |
When we, for each column, add the cells of the first and the second row, we find that the sum is equal. The sum of the last row in the table then becomes n2*(n^{2}+1), since the number of cells are n^{2} and the content of each cell is n^{2}+1. This sum is however equal to 2*S.
The sum of all numbers in the magic square will then be: S = n^{2}*(n^{2}+1)/2.
The magic square consists of n rows. Dividing S with the number of rows (n) gives the sum of each row, i.e. the magic sum will then be equal to n*(n^{2}+1)/2!
Lets check this result with the result in my previous blog post by inserting n=3 in the formula above!
n*(n^{2}+1)/2 = 3*(3^{2}+1)/2 = 15. Yes, the generic formula matches the earlier result!