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New Puzzle

 
 
DrewDad
 
Reply Mon 10 May, 2004 03:30 pm
You have a game with 44 tokens. You and your opponent take turns picking up the tokens, and each may choose to pick up 1, 2, 3, 4, 5, or 6 tokens on his or her turn. The winner of the game is the player that picks up the last (44th) token. Describe the winning strategy.
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Type: Discussion • Score: 0 • Views: 1,597 • Replies: 13
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Tryagain
 
  1  
Reply Tue 11 May, 2004 04:47 pm
Ok, I will have the first go:

The most that can be taken is 6+6. The least 1+1 You start with an even number. You finish with an even number +1. If your opponent chooses odd, you choose odd, (back to even). If they choose, even you do the same.
Your opponent must be left with 8 on their turn for you to be certain to win.

Well at least I tried. Smile
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MyOwnUsername
 
  1  
Reply Wed 12 May, 2004 02:30 am
if you left him with 8 he takes one and you are left with 7. you have to take at least one and you cannot take 7. therefore he wins....

you have to leave your opponent on 7, 14, 21, 28 and 35. So, if you are the one that goes first it's easy - you have to take 2 tokens. He is left with 42. Whatever he takes you leave him on 35. Then whatever he takes you leave him on 28. Whatever he takes you leave him on 21. Whatever he takes you leave him on 14. Whatever he takes you leave him on 7. And then you won.
If he is the one that goes first I don't know if there is sure way for you to win, unless he makes mistake in first move.
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Tryagain
 
  1  
Reply Wed 12 May, 2004 06:31 am
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DrewDad
 
  1  
Reply Thu 13 May, 2004 07:29 am
To win: you can guaranty a win by being the player to end his or her turn by picking up the 2nd, 9th, 16th, 23rd, 30th, or 37th token. At that point, no matter how many tokens your opponent chooses you can force a win.
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MyOwnUsername
 
  1  
Reply Thu 13 May, 2004 03:35 pm
that's what I said just backwards Smile
2nd, 9th, 16th, 23rd, 30th and 37th token equals 42, 35, 28, 21, 14 and 7 remaining
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DrewDad
 
  1  
Reply Mon 17 May, 2004 02:43 pm
I know. Smile
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neitsnie
 
  1  
Reply Tue 25 May, 2004 03:28 pm
Much simpler way, take the first two, then make sure that your opponents move+your move, equal 7; i.e. your opponent takes 4, you take 3. 3+4=7 get it. This strategy is adaptable for all versions of this game. If I remember right it's called Nim; if not, someone please correct me and tell what Nim is, aside from it being a game of course.
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MyOwnUsername
 
  1  
Reply Thu 27 May, 2004 04:49 am
and once again - it is exactly what was said Smile You take 2 (score is 42) and then you took 7 in combination with your opponent so 35, 28, 21, 14, and 7 are left
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magnum
 
  1  
Reply Thu 27 May, 2004 05:28 am
Nim is something else

then you have N number tokens placed in three horizontal rows

each player may remove as much tokens as he like only if 'the tokens are on the horizontal row

like this:

111
1111111111
1111111
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MyOwnUsername
 
  1  
Reply Thu 27 May, 2004 11:45 am
magnum, can you remove tokens from the middle of the row?
for example that it goes like:

111
111 1111
1111111
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billygan
 
  1  
Reply Sat 29 May, 2004 09:48 am
I guess
Im guessing Question
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magnum
 
  1  
Reply Sat 29 May, 2004 07:43 pm
what do you mean mou??

111
1111111111
1111111

and then you remove tokens from the middle row
and then you get

111
111111
1111111


if this is what you mean it's a yes
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MyOwnUsername
 
  1  
Reply Sun 30 May, 2004 01:25 am
no it was not shown as I wanted Smile

I thought if you have

111
1111111111
1111111

and you take from middle of middle row, so you are left with:

111
111----111
111111

or you have to take it from the end?
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