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Fred and George are to share a chocolate bar made up of an 8 by 6 square rectangular array. The top-left top-right squares each contain a visible nut.

They take it in turns, starting with Fred, to snap the chocolate into 2 pieces along a line between 2 rows or 2 columns, & then eat 1 of the 2 pieces.

Neither Fred nor George wants a nut. Which person can guarantee that they won’t get the nut? Is this true for any size of bar?

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closed as off-topic by 900 sit-ups a day, PVAL, rogerl, Sami Ben Romdhane, Goos Aug 3 at 23:59

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1  
I don't get it, are there nuts elsewhere in the chocolate bar? –  muzzlator Feb 24 '13 at 13:20
1  
i dont know either :( –  amy Feb 24 '13 at 13:21
    
I think the question is, if both are trying to avoid the visible nuts...who, under the constraints, will be able to avoid eating a visible nut. –  amWhy Feb 24 '13 at 13:23
    
Do they stop when one eats a nut, or when the entire bar is gone? –  anorton Feb 24 '13 at 13:24
    
i think its till they eat a nut but its only to guarantee who wont? –  amy Feb 24 '13 at 13:26

2 Answers 2

I'm pretty sure it only makes sense to keep the game going until one eats a nut. Here's the strategy for Fred to not eat the nut:

  1. Fred breaks the chocolate bar along the top row (that is, so the row containing the two nuts is the only one remaining). He eats the non-nutty portion.
  2. George must now eat a nut.
  3. Game over. :)

I'll let the reader extend this to other size bars.

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If the question is if it is possible to go on and have the opponent eat both nuts, this can't be done. Each division of the 2-nuts bar leaves either a 2-nut bar and a nutless bar, or two 1-nut bars. The exact way of dividing doesn't matter, as long as it divides the bar into two connected pieces.

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