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The problem refers to a 3x3 magic square using numbers 1-9 without repetition, where each row, column, and diagonal sums to the same value.

Key Observations:

• The sum of numbers 1-9 is (1+2+...+9=45). Since there are 3 rows, each row sum is (45\div3=15).
• The center cell is part of 4 lines (middle row, middle column, two diagonals). Let the center be (C):
  The total sum of these 4 lines is (4\times15=60), which counts the center 4 times and all other cells once.
  So: (4C + (sum\ of\ other\ cells) =60).
  Since the sum of other cells is (45-C), substitute:
  (4C +45 -C =60\ →\ 3C=15\ →\ C=5).

Answer: 5.

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