10.2. Pete and Vasya take turns writing natural numbers on the board, not exceeding 2018 (writing a number that is already on the board is prohibited); Pete starts. If after a player's move, there are three numbers on the board that form an arithmetic progression, this player wins.

![](https://cdn.mathpix.com/cropped/2024_05_06_9562f571445df680906cg-3.jpg?height=309&width=311&top_left_y=415&top_left_x=979)

Fig. 4 Does either player have a strategy that guarantees a win?

(M. Diden, P. Kozhevnikov)
Answer. Vasya.

Solution. Consider the moment after the third move (when three numbers have been written). If no one has won by this point, Vasya wins on the next move - he just needs to find two written numbers of the same parity and write their arithmetic mean (which is an integer).

Moreover, note that if three integers from the set $1,2,3, \ldots, 2018$ form an arithmetic progression, then its difference is no more than 1008 (otherwise, the difference between the largest and smallest numbers would be at least $2 \cdot 1009=2018$, which is impossible).

Now let's describe Vasya's winning strategy.

Suppose Petya's first move is to write the number $a$. Assume that $a \leqslant 1009$. Then Vasya writes the number 2017 or 2018, whose parity is different from the parity of $a$ (denote this number by $b$). After this move, two numbers of different parities are written; therefore, they cannot be the first and third terms of a progression of integers. Since $b-a \geqslant 1009$, they also cannot be consecutive terms of a progression. Thus, Petya
cannot win on the third move. But in this case, as we saw earlier, Vasya wins on the next move.

If $a \geqslant 1010$, then Vasya responds by writing the number 1 or 2, which has a different parity from $a$. The subsequent reasoning is analogous to the first case.

Comment. A correct answer without justification (or with incorrect justification) - 0 points.

Presenting a strategy that does not work in at least one case - 0 points.

A correct strategy for Vasya is presented without justification that it works - 5 points.

It is only proven that at the moment when three numbers are written on the board, the next move always wins - 2 points. 10.3. Positive numbers $x, y$ are such that $x^{5}-y^{3} \geqslant 2 x$. Prove that $x^{3} \geqslant 2 y$.

(N. Agakhanov)

Solution. The required inequality is equivalent to the inequality $x^{9} \geqslant 8 y^{3}$. By the condition, $8\left(x^{5}-2 x\right) \geqslant 8 y^{3}$; thus, it is sufficient to prove the inequality $x^{9} \geqslant 8 x^{5}-16 x$. By moving all terms in the last inequality to the left side, we obtain the inequality $x\left(x^{4}-4\right)^{2} \geqslant 0$, which is true for positive $x$. Therefore, the required inequality is also true.

Comment. The problem is reduced to proving the inequality $x^{9} \geqslant 8 x^{5}-16 x$ (or an equivalent one) for positive $x-3$ points.