You have six balls - three red and three black. All the red balls weigh differently, i.e. one of them is heavy, the other medium, and the third light. Each red ball has a black twin of the same weight. A heavy and a light ball put together weigh as much as two medium balls.
What is the least number of weighings required on a balancing scale to determine which is which?
As there are 36 orderings of the six balls, the theoretical minimum is four weighings. This minimum can in fact be accomplished.
The procedure is to weigh any two red balls and a black ball in 'pan 1' against the other red ball and two black balls in 'pan 2'.
If the scale remains balanced, remove the red ball and one of the black balls from 'pan 2' and transfer one of the red balls from 'pan 1' to 'pan 2', then weigh, noting the results. Then, swap one pair of like colored balls already on the scale, and weigh again, noting the results. Finally weigh the two balls initially removed after the first weighing in one pan against one red and one black (which balls depends entirely upon the results of the two prior weighings). And, then, by deduction, the weight of each ball can be determined.
A similar method can be performed when the scale shows an imbalance after the initial weighing. By weighing selected pairs of colored balls in three subsequent weighings, the weights of each colored ball can be deduced.
Posted by Dej Mar
on 2006-07-11 15:35:41