Two players play the following game. At the outset there are two piles, containing $10,000$ and $20,000$ tokens,respectively . A move consists of removing any positive number of tokens from a single pile $or$ removing $x>0$ tokens from one pile and $y>0$ tokens from the other , where $x+y$ is divisible by $2015$. The player who can not make a move loses. Which player has a winning strategy
Problem
Source: Baltic Way 2015
Tags: winning positions, Combinatorial games, combinatorics
gavrilos
09.11.2015 16:00
Hello. Let $A,B$ be the players,with $A$ playing first.We will show that $A$ has a winning strategy. At his first move $A$ removes a total of $28210=14\cdot 2015$ tokens,$9105$ of which are from the first pile,and $19105$ from the second.The piles are now left with $895$ tokens each.Obviously $B$ can remove tokens from one pile every time since we are left with $1790<2015$ tokens.The rest of the strategy is the following: If $B$ remove $k$ tokens from one pile,then $A$ removes $k$ tokens from the other. Eventually $B$ will empty some of the piles.$A$ then will empty the other and he will win.
rafaello
14.08.2021 00:27
First player has a winning strategy. Strategy is described as follows: First player takes $9105$ tokens from the pile containing $10000$ and takes $19105$ tokens from the pile containing $10000$ as $2015\mid 28120= 9105+19105$. Now both piles contain $895$ tokens. Only moves that can be made are single pile moves as $895+895=1790<2015$, hence first player makes the exact move on other pile, i.e. plays symmetrically. First player makes the last move.