Problem. Magic
Square (click here to see the problem statement)
My Solution
(21 in Center).
If 21 is to be the number in the center
square, we can assume that the sum of each row, column, and diagonal
is the product of 21 and 3: That is the AVERAGE value of the squares
is 21. This establishes a target sum of 63. The sum of the remaining
pairs must equat 63 - 21 = 42.
I make a list of 20 pairs:
|
1, 41 |
2, 40 |
3, 39 |
4, 38 |
5, 37 |
|
6, 36 |
7, 35 |
8, 34 |
9, 33 |
10, 32 |
|
11, 31 |
12, 30 |
13, 29 |
14, 28 |
15, 27 |
|
16, 26 |
17, 25 |
18, 24 |
19, 23 |
20, 22 |
Here are several solutions I found. {I have no doubt there
are many others. A good question would be - How solutions are
there? But this will have to wait for another day}:
