Bloggers Wanted
We're looking for people to help with the main blog. If you are consistent, knowledgeable and you're into it, please drop me a note.
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imported_Bojan
 Senior Boarder
Posts: 78 
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I liked Justin Leck's answer, and I've got another one, too.
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Put 1 tiger in each cage and the 6 or more tigresses anywhere you want. Note that a combination of masculine and feminine objects is a masculine collection.
Jon Miller
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Sweety
Senior Boarder
Posts: 69
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The question asks that we not have 'more than one tiger in a cage.' If we take this to mean that no cage should have more than one tiger, how does your solution help? In your solution, the outer cage has nine tigers in it. After all, intersecting cages, or enclosing cages in other cages doesn't change the fact that we have three cages. (Or, if you take the position that a 'cage' is any enclosure that a tiger cannot leave, then you've violated the condition that we only have three cages.)
It seems to me that the only solution is to read the question strictly. The question asks that we have no 'more than one tiger in
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paydayuscf
Senior Boarder
Posts: 79
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Actually, I think I can get 25 tigers into three intersecting cages. (Or, rather, I can get 25 separate regions (plus the exterior) by intersecting three squares. I could probably get more by intersecting cubical or rectangular solid cages, but the drawings for that are horrible.)
Draw one square.
Draw a second of the same size and centered on the same point as the first, but rotated 30 deg clockwise.
Draw a third of the same size and centered on the same point as the first, but rotated 60 deg clockwise.
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