1. Igor and Pasha are playing a game, taking turns placing natural numbers at the vertices of a regular hexagon (each can choose any free vertex and place any natural number in it). After six moves, when the game ends, the judge writes the product of the numbers at the two ends of each side of the hexagon on each side. Then all 12 numbers are summed. If the sum is odd, Igor wins, and if it is even, Pasha wins.

It is known that Igor moves first. Who of the players can win regardless of the opponent's actions and how should they act?

(S. S. Korechkova)
Answer: Pasha.

Solution. Let's number the vertices in a circle from 1 to 6 and mentally divide them into 3 pairs of diametrically opposite vertices (1-4, 2-5, 3-6).

To win, Pasha should place the same number that Igor placed on his previous move in the opposite vertex. Then, by the end of the game, all numbers in opposite vertices will be the same, so the numbers on opposite edges will also be the same. Thus, the sum of the numbers in the vertices will be even, and the sum of the numbers on the edges will also be even. Therefore, the total sum will be even.

Criteria. Correct answer without justification - 0 points.

If Pasha's incorrect strategy involves symmetry or if a symmetric strategy is incorrectly justified - 3 points.