>>> 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. > > 9 balls (Four normal men and Hitler) > Sorry I misread the question > For this q the answer is 4
I think we can stretch it to five since this guy has at least one defective ball ;-) If the scales dont tilt either way, in the first three tries, we can safely assume the fifth one is defective. And this solution is good enough for balls of all sizes. Lukhman.
