This is our first puzzle..
U hav 12 identical coins out of which 1 has a slightly different weight (It may be heavier or lighter).U r given a balance and u r asked to find out the defective coin using the balance. Wat is the minimum number of trials required to find out the defective coin??
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8 comments:
three
please post how u arrived at the answer..
the solution is very very complicated ..it is 3
Most of you who have got an answer to this would have obtained it by finding a weighing procedure. (4,4,4) and so on. But this question can be answered from information theory alone (though it won't give a weighing procedure).
The total number of possibilities are 24 (any of the 12 coins can be odd and it can be either heavy or light => 12 x 2 = 24 choices)
Each weighing can be smartly done to give an answer to a 3 choice question rather than a 2 choice one.
So number of 3 choice qns required to answer a 24 choice question is n where 3^n >= 24. Hence n = 3.
More details after u complete info theory course.
oh wow.. cant wait to learn info theory now.. :)
The answer is three...
first divide it into two groups of 6...n weigh them...the one containing the heavy one will move down...now take the set containing that and divide it into groups of 3...take any one 2..n one on each side...if any one of them moves down..that is the coin...otherwise..the one outside is our required coin...total no. of moves=3
@arjun --you are an Idiot...u assumed the coin is heavy....
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