Variant 1
You have eight small stone balls, where one is slightly heavier than the others. You have an old-fashioned balancing scale. In as few weighings as possible, how do you find the heavier ball?
Answer
You need two weighings.
First, weigh three balls against three balls. If they balance, the heavier is one of the remaining two, which you can find in the next weighing.
Otherwise, one side was heavier in the first weighing, and one of those three balls must be heavier. For the second weighing, take two of those prospects and weigh them. If they balance, the one left out is heavier. If they don’t balance, one of them is heavier, and it is the answer.
Variant 2
You have twelve small stone balls, where one is of slightly different weight than the others, either lighter or heavier - you don’t know. You have an old-fashioned balancing scale. In as few weighings as possible, how do you find the odd ball?
Answer
You need three weighings.
In the first weighing, weigh four balls against four balls.
If they balance, the odd ball must be one of the remaining four balls, and the eight balls in the first weighing are all of regular weight. For the second weighing, take two unknown balls and weigh them against two balls whose weights are known. If they balance, the odd ball is one of the other two balls. If they don’t balance, one of the unknown balls on the scale is odd. For the third weighing, take one of the two unknown balls and weigh it against a regular ball. If they balance, the one left out is odd, otherwise it is the one on the scale.
If the first weighing didn’t balance, we have eight possibly odd balls and four regular ones. In the first weighing, one side was lower, and one side was higher, resulting in four balls that are possibly heavier and four balls that are possibly lighter than the others.
For the second weighing, place two possibly-heavier balls and one possibly-lighter ball on the left side, and two possibly-heavier balls and one possibly-lighter ball on the right side.
If the scale balances, one of the remaining two unknown balls - two possibly-lighter balls - is odd. Take one of them and balance it against a known ball, and the answer is given.
If the scale didn’t balance in the second weighing, then either the left side is lower than the right side, or vice versa. If the left side is lower, then one of the two possibly-heavier balls on the left side or the possibly-lighter on the right side is odd. And if the right side is lower, then one of the two possibly-heavier on the right side or the possibly-lighter on the left side is odd.
In either case we have two possibly-heavier balls and one possibly-lighter balls remaining that could be odd. For the third weighing, take the possibly-lighter ball and one possibly-heavier ball and place them on the left side, and two regular balls on the right side. If the scale balances, the ball left out is odd. If the scale doesn’t balance, then the possibly-heavier ball is odd if the left side is lower than the right side, and the possibly-lighter ball is odd if the left side is higher.
(If anyone knows of a better way to present the answer to this puzzle, please let me know. Surely this can’t be the best way to do it…)