Expert: Sherman D. - 11/4/2005

Question
The question is as follows:

For this problem you have a number of metal balls whick all look and feel identical. One of the balls weighs slightly than the others ,although not enough for you to notice the diffrence without mechanical help.

All you have to find the heavier metal ball is a set of old fashioned scales that can only tell you whichtray is heavier than the other. you can do a test of weight with any number of balls in each of the trays.

Your task is to find WITH CERTAINTY the heaviest ball.

Show, WITH REASONING ON HOW IT IS DONE,the minimum tests required to find the heavier ball if you have 2,3,4,5,6,7,8,9 or10 balls in total.

Show, again with reasoning, the minimum tests required to find the heaviest ball out of 100 balls.


Answer
I will number the balls form 1-10

2 balls - 1 : 2 = one try

3 balls - 1 : 2 = 1 try, since if 1 and 2 weighs the same, that leaves 3 as the one that weighs the most, if that isn't the case, then either 1 or 2 is the one that weighs the most.

4 balls - 1,2 : 3,4 to 1 : 2 = 2 tries

5 balls - 1,2 : 3,4 = 1 or 2 tries, same reason that i had for 3 balls

6 - 1,2,3 : 4,5,6, to 1 : 2 = 2 tries

7 - 1,2,3 : 4,5,6 to 1 : 2 = 1 or 2 tries

8 - 1,2,3,4 : 5,6,7,8 to 1,2 : 3,4 to 1 : 2 = 3 tries

9 - 1,2,3,4 : 5,6,7,8 to 1,2 : 3,4 to 1,2 = 1 or 3 tries

10 - 1,2,3,4,5 : 6,7,8,9,10 to 1,2 : 3,4 to 1 : 2 = 2 to 3 tries

The minimum test is 1 for odd number of balls and 2 for even number of balls, however, this is only the mininmum test when dealing with probability. If you are lucky enough to get both sides equal leaving one ball left over, then that would be a probability chance.

So i can't be sure on the precise answer for mininmum tests since the answer varies on probability.