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Topic: Jewel Thieves (Read 2091 times) |
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Garzahd
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Jewel Thieves
« on: Jan 29th, 2003, 12:31am » |
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Two thieves conspire to steal a valuable necklace made of diamonds and rubies (evenly spaced, but not necessarily alternating or symmetric). After they take it home, they decide that the only way to divide the booty fairly is to physically cut the necklace in half.
Prove that, if there is an even number of diamonds and an even number of rubies, it's possible to cut the necklace into two pieces, each of which contains half the diamonds and half the rubies.
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| « Last Edit: Oct 23rd, 2003, 8:04pm by Icarus » |
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Speaker
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Re: New puzzle: Jewel Thieves
« Reply #1 on: Jan 29th, 2003, 1:27am » |
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Probably this isn't what you are expecting, but it satisfies my high standards of answerness.
Slice the necklace across a plane such that each of the jewels is cut in half, thus providing the same number of rubies and diamonds on each half. But, no string.
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Garzahd
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Re: New puzzle: Jewel Thieves
« Reply #2 on: Jan 29th, 2003, 9:40am » |
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The thieves came up with that idea, but they thought it would be too much work to steal diamond cutting tools along with the necklace. Plus, cutting each gem in half would reduce its value more than just cutting the whole necklace in two places.
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Phil
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Re: New puzzle: Jewel Thieves
« Reply #3 on: Jan 29th, 2003, 10:33am » |
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Assuming the necklace is an unbroken circle to start with, not a string with a clasp, you must divide it at two opposite points. If one side has more diamonds than the other, you rotate the necklace one jewel. If the switched jewels were the same, it changes nothing and you rotate again. If they're different, the number of diamonds on the left half will either go up one or down one. Eventually, by the time you've rotated the necklace 180 degrees, the left side will have the number of diamonds the right side used to have. Since each rotation changes the number by one, and there's an even number of diamonds, at some point they must have been equal. If the diamonds are equal then the rubies are also equal since each half has the same number of jewels.
// hidden by moderator 11:50 PM 1/30/2003
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| « Last Edit: Jan 30th, 2003, 11:50pm by william wu » |
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Garzahd
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Re: New puzzle: Jewel Thieves
« Reply #4 on: Jan 29th, 2003, 2:46pm » |
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Nicely done! Have you seen this one before, or did you figure it out?
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