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Weighing in a Harder Way
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.
Posted 22 January 2010 - 07:29 PM
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.
Posted 22 January 2010 - 11:30 PM
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..
Posted 27 May 2010 - 07:21 AM
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
Posted 13 July 2010 - 04:27 AM
An easier-to-understand (IMHO): (sorry for my bad english)
Edited by Teddy Kesgar, 13 July 2010 - 04:30 AM.
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 ..
Posted 14 December 2010 - 06:48 AM
overall 4 weighing...
Posted 18 December 2010 - 09:10 PM
Posted 26 December 2010 - 11:52 AM
Posted 20 January 2011 - 12:38 PM
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