All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Games
Who will win? (Posted on 2005-01-10) Difficulty: 4 of 5
There are 2n cards labeled 1, 2, ..., 2n respectively, and the cards are distributed randomly between two players so that each has n cards. Each player takes turns to place one card, and you win if you put down a card so that the current sum of all the played cards is divisible by 2n+1.

For example, if n=10, and the previously placed cards are 5, 8, 9, 19, then if player A now places 1, he wins since 5+8+9+19+1 = 42 is divisible by 2*10+1=21.

Assuming both players want to win, what strategy should one adopt in order to win? Following the strategy, is there a consistent winner of this game?

See The Solution Submitted by Bon    
Rating: 2.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Some Thoughts re: Solution | Comment 2 of 9 |
(In reply to Solution by David Shin)

I can see that, if the second player gets to play his last card, he will win, since the total sum is n(2n+1) a multiple of (2n+1).

But, why is it that only k choices will lead to a win by the first player?
  Posted by e.g. on 2005-01-10 18:07:22

Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information