20051117, 20:08  #1 
2^{3}×3×11×31 Posts 
100 wooden sticks
Ok, I need help solving this problem...
Players A and B play a game involving 100 wooden sticks that are piled between them. They are to remove from one to six sticks during each turn. The player who removes the last stick is the winner. Player A goes first. Is there a way for her to force the outcome so that she will win? If so, what should be her strategy? Thanks for any help! 
20051117, 20:45  #2  
Nov 2003
2^{2}×5×373 Posts 
Quote:
Hint: Think "0 mod 7" 

20051117, 21:24  #3  
Oct 2004
23^{2} Posts 
Quote:


20051118, 06:29  #4  
Jun 2005
Near Beetlegeuse
2^{2}×97 Posts 
Quote:
A removes 2 sticks on her first go. However many sticks B removes, A ensures that the number of sticks remaining is a multiple of 7. When there are only 7 sticks remaining and B has to go he can only take > 0, < 7 sticks and so A removes the remaining sticks and wins. 

20051118, 07:12  #5  
Oct 2004
23^{2} Posts 
Quote:
(which was brilliantly concise, understated and required THOUGHT to apply the solution given) I was simply pointing out that if one was in posession of the solution (as we now all are) but was NOT the first player to make a move, you could indeed know whether the first player took the opportunity to make the perfect play move or gave that opportunity away, in which case it would be yours for the taking. In other words: Although A can force a win by perfect play, it is also true that if A deviates even once from perfect play, then B can force a win by perfect play. Let's assume you were someone who knew the technique (you do now) and let's assume your opponent didn't (this should be quickly apparent). It is unlikely your opponents would be willing to let you go first every time as they would consider this an unfair advantage. Thus in repeated matches you are likely to alternate as first/second player to move. Alternatively you could toss a coin to determine the starting player which may make it harder for them to realise the pattern that your winning often has something to do with going first. In those games you are first you will always force the win (unless to give the opponent false confidence and increase gambling stakes prior to wiping them out). In games where you are second, but the opponent slips up you can turn these around in your favour. Of course you can further hide your strategy by deliberately throwing some games where you could have won so that it appears you do not always win games you start. Cunning plan eh? If your opponent also knows the technique but is letting you win, it is a very hilarious double bluffing situation where each of you is trying to determine how smart the other is (rather like bluffing a poker hand). I imagine this simple maths could be used to trick children out of sweets, and in states where gambling is legal, to con foolish adults out of real money! Some people might view that as exploitation. Last fiddled with by Peter Nelson on 20051118 at 07:25 

20051120, 17:33  #6  
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}×3^{3}×19 Posts 
100 wooden sticks
Quote:
Numbers: Thanks for the explanation to the hint. I was lost for awhile without it. Nelson: What you say follows logically. Mally 

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