Let $n$ be a positive integer, the players A and B play the following game: we have $n$ balls with the numbers of $1, 2, 3, 4,...., n$ this balls will be in two boxes with the symbols $\prod$ and $\sum$. In your turn, the player can choose one ball and the player will put this ball in some box, in the final all the balls of the box $\prod$ are multiplied and we will get a number $P$, after this all the balls of the box $\sum$ are added up and we will get a number $Q$(if the box $\prod$ is empty $P = 1$, if the box $\sum$ is empty $Q = 0$). The player(s) play alternately, player A starts, if $P + Q$ is even player A wins, otherwise player B wins. a)If $n= 6$, which player has the winning strategy??? b)If $n = 2012$, which player has the winning strategy???
Problem
Source: Olympiad
Tags: combinatorics, number theory
23.01.2018 17:07
23.01.2018 17:47
Anybody trying this........... Can people please verify wether my answer is correct or not ?
23.01.2018 18:42
...........
23.01.2018 18:43
How, can you tell me the strategy.............
23.01.2018 18:44
Yes wait.
23.01.2018 18:50
............
23.01.2018 18:55
I have a different strategy for $B$, denote the numbers $1,2,3,.2012$ by $1,0,1,0,1....,0$ ( Only odd and even matters $0$ for even and $1$ for odd) Then, Claim 1: The one to put last $1$ is the winner.( To prove this, look at cases like box 1 is even and the second is odd etc..) For $n=4$, we claim B is winner..... ( it's trivial). Hence for $n$ is even, B can always get the remaining numbers to ${0,1,0,1}$ and then $B$ can always choose the last $1$, hence for any even B can win.........
24.01.2018 03:53
Another simple observation If the last person take the odd number, he always wins the game. Hence if n is odd, then A wins. If n is even then A must take odd every time. So B's strategy is to put even number in $\Sigma $ every time except for the last one. In the last selection he needs to consider the situation and puts the number and wins the game.
24.01.2018 05:14
ltf0501 wrote: Another simple observation If the last person take the odd number, he always wins the game. Hence if n is odd, then A wins. If n is even then A must take odd every time. So B's strategy is to put even number in $\Sigma $ every time except for the last one. In the last selection he needs to consider the situation and puts the number and wins the game. That's what my solution was......