On Thursday 10 Sep 2009 10:19:44 am lukhman_khan wrote: > > Note that I am looking for an ability to solve > > such problems in a candidate. > > Can you plz rework the solution and tell me this - > > A maximum of how many spheres can you weigh against each other (only one on > each side) to find the defective one in a max of 3 tries. > > Lukhman.
9 balls (Four normal men and Hitler) Once you have it down to 3 balls and you know whether you are looking for heavier or lighter, 3 balls is the max you should be left with to figure it out in ONE weighing That means that you have only two tries left to find if the defective ball is heavier or lighter Let me explain: Suppose you start with 9 - you put three a side and one side goes down. Then remove the HEAVIER balls and replace with unweighed 3 balls. If equal - your defective ball is HEAVIER and present among the three you just removed. However if the new three balls are also heavier than the original "lighter" side - it means that the defective ball is LIGHTER and present among the three you have as "lighter" You have now used up 2 of three weighings Now take any 2 of the 3 balls containing the defective ball. If equal the third ball is defective. if unequal - you already know whether the defective ball is lighetr or heavier so pick that one. shiv
