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Weighing in a Harder Way


Best Answer bonanova, 18 January 2008 - 06:49 PM

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.

Spoiler for Here's a very slightly modified approach
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#1 wangsacl

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Posted 18 June 2007 - 05:03 AM

You've got 27 coin, each of them is 10 g, except for 1. The 1 different coins has a masses different. It could be 9 g, or 11 g (heavier, or lighter by 1 g).
The question is, in how many minimum ways, you can determine the different coins, and is it heavier, or lighter?
I have no answer to this question.
Do you?
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#2 mdsl

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Posted 18 June 2007 - 01:53 PM

Assuming you have a simple scale, my answer is 9.

First divide the coins into 3 piles of 9.

Measure 2 piles at a time until you find the pair that is equal in mass. This can take up to 3 tries, but if you were lucky you could get it on the first try.

Put the piles of coins with equal mass to the side.

Repeat the process this time with 3 piles of 3 coins, maximum 3 measurements.

Repeat the process this time with the 3 remaining coins, find the two that are equal, the remaining coin has a different mass.

If you are a very luck person you coulp potentially only have to make 4 meaurements, 3 to rule out equal pairs and 1 to compare the final coin to the rest.

9 is the most you would have to do the efficient way, and at any point you can observe is the unequal pile has more or less mass.
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#3 CedricLi

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Posted 16 January 2008 - 12:28 PM

I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it?
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#4 bonanova

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Posted 16 January 2008 - 05:14 PM

I guess the question is lack of some information such as, by using a scale with weight of 1g, 2g, 5g, and 10g, isn't it?

It's a balance scale that compares what's in the two pans.
You can get the answer by just comparing groups of coins.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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#5 bonanova

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Posted 16 January 2008 - 07:21 PM

Duplicate post.
Looking around the forum for an eraser ... <_<
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
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#6 PDR

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Posted 18 January 2008 - 01:05 AM

Spoiler for fewer than 9...

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#7 PDR

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Posted 18 January 2008 - 01:05 AM

Spoiler for fewer than 9...

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#8 PDR

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Posted 18 January 2008 - 01:28 AM

sorry - accidentally hit submit - here's my answer...
Spoiler for fewer than 9...

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#9 Jkyle1980

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Posted 18 January 2008 - 04:24 AM

I think I've got it in 4 measurements:

Separate the coins into 3 stacks of 9 (A, B, C). Weigh stack A against B and then A against C. Take the stack with the different weight (note lighter or heavier) and break it into 3 stacks of 3 (D, E, F). Weigh stack D against E. If D and E are equal, then F is the odd stack. If D and E are not equal, the lighter or heavier (based on the A, B, C comparison) is the odd stack. You now have three coins (G, H, I). Weigh G and H. If G equals H, then I is the odd and is lighter or heavier (based on the A, B, C comparison). If G and H are not equal, then the lighter or heavier (based on the A, B, C comparison) is the odd coin.
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#10 bonanova

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Posted 18 January 2008 - 06:49 PM   Best Answer

Jkyle1980's solution is correct, and much less verbose than what follows.

You can distinguish among 3**N cases in N weighings.
There are 54 possible cases in this puzzle [one of 27 coins is heavy or light].
So three weighings [27 cases] won't do it, but four [81 cases] can.
Spoiler for Here's a very slightly modified approach

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell




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