HINT FOR PUZZLE-4
One of the methods to approach the problem is to just keep trying several permutations and combinations and hope to hit the jackpot. Of course, many people may instinctively discard the obviously impossible alternatives as sub optimal.
Another way, which, I think, having a high chance of getting the most optimal solution is to understand the most inefficient method and slowly improve upon it, step by step, to see where the optimum solution may hiding itself.
Weigh one at a time:
The least mentally straining way to identify the odd ball is to select any ball as a standard and weigh all the other 11 balls one after the other against it. In this method, you may locate the odd ball on the second weighing itself if you are lucky; if you are extremely unlucky, you may have to go through all the 11 weighing.
Weigh two at a time:
The next best method one can think of is to segregate the 12 balls into 6 lots of 2 balls each. Pick any one such lot as a standard and weigh all the other 5 lots one after the other against it. Of course, since the ‘standard lot’ as well as the ‘ test lot’ has two balls, after isolating the ‘odd lot’, you may need to do one more weighing by breaking the odd lot and weighing any one with a known good ball.
In this method, if you are extremely lucky, you will need 3 weighing; if extremely unlucky you will need 6 weighing. (You may initially think that 7th weighing is required to isolate the odd one. But it is not so. As you come to the last weighing, at the end of 5th weighing itself you will realize that the odd ball is in the untested 6th lot; the intelligent thing to do is to break up the last lot and weigh any one ball with a known good ball.)
Thus you will notice that an interesting thing has happened. Taking a test lot as two instead as one, reduces the maximum required weighing by nearly half- 11 to 6; but simultaneously increases the bare minimum required from 2 to 3.
We can
go on discussing like this of choosing test lots as 4 lots of 3, 3 lots of 4
and 2 lots of 6. The total number of balls has been chosen as 12, because it
has got as many as 5 relevant divisors-1,2,3,4 and 6. {Though 12 is also a divisor, it is irrelevant as if
you have a test lot of 12, what will you check them up with?}
We will
stop our discussion here as we may drift slowly towards the solution itself.
Happy hunting!
