Suppose first that in a line of three numbers, adjacent numbers have the same difference. Let n be the first of these numbers, and d be the common difference. Then these numbers are n, n + d and n + 2d. The sum of these numbers is 3n+ 3d and therefore is a multiple of 3.

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Adjacent numbers in each row have a common difference 1. Therefore, the sum of the numbers in three adjacent cells in the same row is always a multiple of 3.
There are two lines of three adjacent cells in each row, for example 1,2,3 and 2,3,4 in the top row.

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Therefore, in the 4 rows there are 4 × 2 = 8 lines of three adjacent cells such that the sum of the numbers in these cells is a multiple of 3.
Adjacent numbers in each column differ by 4. Hence, it follows similarly, that there are 8 lines of three adjacent cells in the same column such that the sum of the numbers in these cells is a multiple of 3.

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We now consider the diagonals from top left to bottom right. Adjacent numbers in these diagonals each column differ by 5. Therefore the sum of numbers in three adjacent cells in the same diagonal is always a multiple of 3.

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One of these diagonals contains 4 numbers. There are 2 lines of three adjacent cells in this diagonal whose sum is a multiple of 3. There are 2 of these diagonals containing three numbers. Therefore, altogether, there are 4 lines of three adjacent numbers on theses diagonals whose sum is a multiple of 3.
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Similarly, on the diagonals from top right to bottom left adjacent numbers have a common difference 3, and therefore there are 4 lines of three adjacent numbers on these diagonals whose sum is a multiple of 3.
Hence, in total, there are 8 +8 + 4 + 4 = 24 lines of three adjacent cells whose sum is a multiple of 3.

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