On Tuesday 08 Sep 2009 7:16:10 pm Kiran K Karthikeyan wrote: > The most efficient use of the first weighing is putting 4 against 4 since > it conclusively eliminates at least 4 no matter what the scenario i.e. if > the pans stay level or if they don't.
This is wrong Six is equally efficient. Whether you check 4 or 6 you can arrive at an exact answer in 4 weighings. This is because whether you choose 4 or 6 it takes a minimum of two weighings to "fix" the defective ball as being lighter or heavier and narrow its position down to a group of 4 or 6. It takes a minimum of two weighings to find a defective ball once you have narrowed your number down to six balls (or four) and you already know that it is heavier or lighter. Once you are left with three (or two) balls, and you know if the defective one is heavier or lighter, you can find the defective ball in just one weighing. However if you click on "I'm feeling lucky" it is possible to get the answer in just two weighings provided you just choose two random balls and one of them happens to be defective. So the asssertion that "It is obviously not possible to do it in less than four" is also wrong. It took me overnight - and about two hours of mental reasoning to arrive at this conclusion shiv
