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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
Go to the full post


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#61 achia

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Posted 22 January 2010 - 07:29 PM

"Parik" posted on 18 March 2009 - 04:46 PM the clearest description I have seen to solving the case comprehensively.

That is by dividing into groups of 9, then 3 and finally singly.
In each Division we can identify the defective group or item in 1 or 2 steps, depending on luck.

Using this method, the best outcome is 3 weighings and the worst outcome is 6 weighings. I think this is the best strategy to adopt as the arithmetic average of the number of attempts is 4.5.

Other strategies like balancing 13-13 and leaving one aside may solve the problem in 1 step (very lucky) or end up with 26 steps (if by 26 attempts the odd coin has not been identified, we can infer that the last coin is the odd one without further weighing). The arithmetic average of all possible outcomes is 13.5.
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#62 aniketgoel

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Posted 22 January 2010 - 11:30 PM

For finding the different one out of three u will always require just 2 attempts..

suppose u try with 1 and 2 bunch they are not equal.

try with 1 and 3 and now if they are also not equal then its sure that the bunch 1 contains fault else if 1 and 3 are equal then bunch 2 will have fault in it.

So just 2 try for a lot of 3.

Thus in just 6 attempts you could find the solution.

So 6 is the answer..
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#63 hareesh

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Posted 27 May 2010 - 07:21 AM

In min 3 weighings and max 5 weighings we can definitely find the odd coin

All possibilities :-
ivide 27 into piles of 7, 7, 13
1) compare 7 vs 7 coins, keeping 13 coins aside
If both of them don't weigh same, then it means that the odd coin is inside these 14 coins
If both of the sets weigh same, then we need to find the odd coin from 13 coins.

2) compare 4 vs 4 coins, keeping 5 or 6 coins aside(this depends on the previous result)
Applying the same logic as explained in step 1, we will be left with 5 or 6 or 8 coins to find the odd coin from

3) Depending on the result from previous step we need to split coins to weigh
case 1) 5 - 2vs2, 1 kept aside
case 2) 6 - 2vs2, 2 kept aside
case 3) 8 - 3vs3, 2 kept aside
If its case one, then we are done, else we will be left with either 2 or 3 coins to determine the odd coin from

4) If two coins are left, then we can find the odd one by comparing with normal coins(other coins)
If three coins are left, then we may require one more weighing
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#64 Teddy Kesgar

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Posted 13 July 2010 - 04:27 AM


An easier-to-understand (IMHO): (sorry for my bad english)
Spoiler for Hint


Edited by Teddy Kesgar, 13 July 2010 - 04:30 AM.

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#65 scorp

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Posted 10 August 2010 - 01:42 PM


An easier-to-understand (IMHO): (sorry for my bad english)

Spoiler for Hint



which one among A or B will you choose for ur next step as it says one of them is either heavy or light ... So u cannot determine the defective bunch this way ..
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#66 jensun

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Posted 14 December 2010 - 02:08 AM

Spoiler for hint

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#67 KlueMaster

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Posted 14 December 2010 - 06:48 AM

Spoiler for Here is how to do it in

overall 4 weighing...
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#68 welshdragon

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Posted 18 December 2010 - 09:10 PM

Using your method you need only 6 weighing .. use a balance scale not a numerical one . with 3 piles you only need to weigh twice, to find the odd one out. weigh pile 1 against 2,. if equal then pile 3 is the odd one . if unequal then weigh the heavier against pile 3 .. this will be equal or lighter . if unequal and it cannot be heavier . that would mean 3 different weights, so pile 3 is the odd ball. do this with the next 2 sets . equaling 6 at most.
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#69 Dumb And Stupid

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Posted 26 December 2010 - 11:52 AM

The answer is 4 and my solution is below.

Spoiler for Proof of Concept

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#70 stalingino

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Posted 20 January 2011 - 12:38 PM

Spoiler for I can make it simple.. ans: 4 times, How..

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