[ Examples and counterexamples. Constructions ]

In the cells of a $3 \times 3$ table, the digits from 1 to 9 were placed. Then, the sums of the digits in each row were found. What is the maximum number of these sums that can be perfect squares?

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The smallest possible sum of digits in a row is $1+2+3=6$, and the largest is $7+8+9=24$. Therefore, if the sum is a perfect square, it must be 9 or 16. Suppose that in each row, the sum is 9 or 16, which means it gives a remainder of 2 when divided by 7. Then the sum of all digits in the table, when divided by 7, will give a remainder of 6. However, the sum of the numbers from 1 to 9 is 45, which gives a remainder of 3 when divided by 7. Therefore, it is impossible to have three such sums.

Two such sums are possible, for example, see the table:

![](https://cdn.mathpix.com/cropped/2024_05_06_43e7ff204cbd87ef9854g-17.jpg?height=263&width=318&top_left_y=2321&top_left_x=870)

## Answer

Two sums.

## [ Arithmetic Progression ]

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