What is the maximum number of natural numbers not exceeding 2016 that can be marked so that the product of any two marked numbers is a perfect square?

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Since $44^{2}1$, we will get a set of fewer numbers, as $1936 q>2016$.

## Answer

44 numbers.

\section*{Problem 65909 Topics:

![](https://cdn.mathpix.com/cropped/2024_05_06_864d5071b377f8fb364bg-02.jpg?height=111&width=729&top_left_y=0&top_left_x=454)

Germann and Chekalin laid out 13 different cards on the table. Each card can lie in one of two positions: face down or face up. The players must take turns flipping one card. The player who, after their move, repeats any of the previous situations (including the initial one) loses. Chekalin made the first move. Who can win regardless of how the opponent plays?

## Solution

Chekalin's winning strategy is to flip the same card (for example, the Queen of Spades) every time. All possible positions can be divided into pairs that differ only by the position of the Queen of Spades. If, in response to Chekalin's move, Germann also flips the Queen of Spades, the previous position will repeat, and he will lose. Therefore, he is forced to flip another card. By flipping the Queen of Spades in response, Chekalin will get a position that is paired with the one that just was. Thus, each move by Germann will have to "start" a new pair, and Chekalin will always be able to make a response move, "finishing" the pair. Since the number of possible positions is finite, sooner or later Germann will not be able to open a new pair and will lose.

## Answer

Chekalin.

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