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Topic: 10 coins and a scale (Read 469 times) |
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You have 10 ancient gold coins and an old fashioned scale. The scale is the kind where you compare weights by putting objects into two different baskets. You know that one of the coins is counterfeit, and that the counterfeit one weighs a different amount than the other coins. What is the minimum number of weighings it will take to tell which coin is the fake one.
Extra info to clarify things:
Each basket can hold up to 5 coins.
Each wieghing tells you which basket full of coins (A or B) is heavier or if they are equal.
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More additional info:
You don't know if the counterfeit coin is heavier, or lighter. Only that is weighs a different amount.
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towr
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Re: 10 coins and a scale
« Reply #2 on: Nov 17th, 2003, 2:34pm » |
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There's allready multiple variants of this puzzle on the forum..
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jtrook
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Re: 10 coins and a scale
« Reply #3 on: Nov 20th, 2003, 9:39am » |
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Where are the other variants listed?
I think this is an interesting puzzle.
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towr
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Re: 10 coins and a scale
« Reply #4 on: Nov 20th, 2003, 11:26am » |
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I would've listed them last time, but the search function is often acting wonky for me lately..
Anyway, without further ado, at least not very much..
Here are the ones I could find:
12 BALLS
12 balls - variation
12 balls variation
Billiard balls
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| « Last Edit: Nov 20th, 2003, 2:00pm by towr » |
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phobos
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Re: 10 coins and a scale
« Reply #5 on: Nov 28th, 2003, 7:50am » |
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Recently I'd been asked a 13-ball version. The conditions remain the same, just that there're 13 balls now. Is it still possible to get it with 3 weightings?
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LZJ
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Re: 10 coins and a scale
« Reply #6 on: Nov 28th, 2003, 8:42am » |
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I believe it is possible.
Split the 13 balls into 3 groups of 4 with the last ball set aside temporarily. Let us call the balls AAAA, BBBB, CCCD and E.
Weigh AAAA against BBBB. If the ball is amongst them (scales not balanced), just chuck D away and continue with the 12 ball variation.
If they are evenly balanced, weigh AAA against CCC. If it is amongst CCC, you'll be able to find out whether the ball is heavier or lighter, and you can use the last measurement to find the ball amongst the CCC.
If it isn't among CCC, it is either D or E. Weigh A against D. If it balances, the different ball is E. If not, its D.
Note, however, that if the ball is E, you will not be able to find whether it is heavier or lighter than the norm using just 3 weighings, but hey, at least it is identified.
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