im stuck with this magic square2 where the sum has to equal 35
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If the sum is 35 for each row, column and diagonal, the numbers must total 3*35=105.

Now we need to find out if the numbers are consecutive. The numbers 1-9, for example, add up to 45.

If the numbers form an arithmetic progression (AP): a, a+d, a+2d, ..., a+8d, the sum is 9a+d(0+1+2+...+8)=9a+d*8*9/2=9a+36d=105; a+4d=11+2/3. Since a and d are supposed to be integers, this equation cannot be satisfied because of the fraction 2/3.

The numbers 1 4 6 9 12 14 17 20 22 could be used to construct a 3x3 magic square (almost).

20  1 14

 6 12 17 all but one sum is 35

 9 22   4

The following numbers form a true magic square:

1 11/3 19/3 9 35/3 43/3 17 59/3 67/3:

59/3      1 43/3

19/3 35/3    17

     9 67/3 11/3

The numbers 5 20/3 25/3 10 35/3 40/3 15 50/3 55/3 similarly form a true magic square.

by Top Rated User (1.1m points)

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