Pasha and Vova play the game crossing out the cells of the $3\times 101$ board by turns. At the start, the central cell is crossed out. By one move the player chooses the diagonal (there can be $1, 2$ or $3$ cells in the diagonal) and crosses out cells of this diagonal which are still uncrossed. At least one new cell must be crossed out by any player's move. Pasha begins, the one who can not make any move loses. Who has a winning strategy?
Problem
Source: Caucasus MO 2023
Tags: combinatorics
crezk
19.07.2023 18:51
I didn't dwell on it much, but for $k\geq2$, $4(k-1)$ moves seem to be made in the most probable game in a $3\times (2k+1)$ table. Since the number of moves will be even, the $2$nd player wins the game.
FredAlexander
13.02.2024 17:57
Im not sure but if first player crosses a 2 cell diagonal. And then if 2nd crosses 1 cell 1st crosses 2 cell, if 1 then 2, if 3 then 3 to make the number divisible by 3
turann
29.08.2024 17:00
FredAlexander wrote: Im not sure but if first player crosses a 2 cell diagonal. And then if 2nd crosses 1 cell 1st crosses 2 cell, if 1 then 2, if 3 then 3 to make the number divisible by 3 Həri
EmptyMachine
26.12.2024 17:43
I guess that Vova has a winning strategy for this one