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Topic: Coin weghting: 13 coins with 2 defected coins (Read 1241 times) |
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wonderful
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Coin weghting: 13 coins with 2 defected coins
« on: Apr 3rd, 2008, 4:45pm » |
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I posted the same question in the "easy" section a a follow-up to Hippo's correct answer for the case of 8 coins. That said, it might be more relevant to post this question here:
There are 13 coins of which 11 weighs A g each. The other two are defected. One weighs (A+ 2) g and the other weighs (A-2) g.
Can you find out the two different coins using minimum weightings?
Have A Great Day!
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| « Last Edit: Apr 3rd, 2008, 7:23pm by wonderful » |
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Grimbal
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Re: Coin weghting: 13 coins with 2 defected
« Reply #1 on: Apr 4th, 2008, 6:00am » |
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Are these weightings on a 2-pan scale or on a scale that gives the weight?
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Hippo
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Re: Coin weghting: 13 coins with 2 defected
« Reply #2 on: Apr 4th, 2008, 1:11pm » |
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I expect its the same puzzle with 8 replaced by 13, so the 2pan scales are expected.
So for the start there is 34<13*12<35 initial positions so we are trying to find solution on 5 measurements.
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| « Last Edit: Apr 4th, 2008, 1:11pm by Hippo » |
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wonderful
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Re: Coin weghting: 13 coins with 2 defected coins
« Reply #3 on: Apr 4th, 2008, 9:14pm » |
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5 is the correct answer. Next, we would like to design a weighting scheme to achieve this target. Result from the 8 coins can be used here.
Have A Great Day!
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wonderful
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Re: Coin weghting: 13 coins with 2 defected coins
« Reply #4 on: Apr 14th, 2008, 2:31am » |
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It has been a while and it seems that nobody has come up with a solution. As such, I would like to give some hints here:
In the first three weighting we can do something like this:
(5, 6, 7,  ? (9,10,11,12)
(2, 8, 11, 12) ? (3, 4, 9, 10)
(1, 4, 6, 10) ? (3, 7, 9, 12)
After that, there are two more weightings.
Let's continue.
Have A Great Day!
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| « Last Edit: Apr 14th, 2008, 2:34am by wonderful » |
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