A 3x3 square:
2) There is no more square to the top right diagonal of square 1, therefore, you put the next number, "2" to the right column last row.
3) To put "3", again, there is no more square to the top right diagonal of 2, therefore, you put it to the row above "2" at the most left column.
5) There is an empty square to the top right diagonal of "4", therefore, fill "5" into the square, followed by "6" to the top right diagonal of "5".
6) To enter "7", there is no square to the top right diagonal of "6" and also no column to the right of "6". Therefore, you fill "7" in the square below
7) To enter "8", there is no square to the top right diagonal of "7", but there is a row above "7", therefore, you enter "8" in the row above "7", and to the most left column.
8) To enter the last number "9", again there is no square to the top right diagonal of "8", therefore "9" is entered in the column to the right of "8" but to the last row.
Do you get it now? In a 3x3 square, every column, row, and diagonal adds up to 15.
This formula can be used for all odd-numbered squares, e.g., 5x5, 7x7, 9x9, and so on.
Try solving the 5x5 square.
Below is the answer for the 5x5 square.
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