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?
The number I come up with is 3 weighings which seems to differ from some of the others posted. This method doesn't take into account every random combination of balls because we are only care about finding the combination which gives the minimum number of weighings.
1. The first step is to weigh 2 red balls and 1 black ball in one pan with the remaining red ball and 2 black balls in the other pan. If they weigh the same then the 1 black ball and 1 red ball from each pan must be the same weight.
2. Then take the 2 red balls and weigh them against the 1 black ball and 1 red ball from the previous measurement. If it balances then the 1 black ball and 1 red ball must be medium and the 2 red balls must be 1 heavy and 1 light. Without measuring the 2 black balls from measurement #1, one can conclude that there is 1 heavy and 1 light.
3. Then take the 2 red balls and 2 black balls from measurement #1 and weigh them as 1 red ball/1 black ball in one pan and 1 red ball/1 black ball in the other pan. If the scale tips showing pan #1 is heavier then the balls in pan #1 are heavy and the balls in pan #2 are light.
A different solution
