I got Trouble in River City. I have to solve every riddle I see just to prove to myself I’m not getting dumber.
I figured it out using pennies as normal balls, dimes as potentially light balls, and nickels as potentially heavy balls.
Step 1: Put 4 balls on each side of the scale.
Step 1 possible outcome: The sides are even. Therefore if there is an odd ball it must be one of the 4 balls you haven’t balanced yet.
Step 2: Replace one of the 4 untested balls with a normal ball. Put 2 untested balls on one side of the balance, and one normal and one untested ball on the other.
Step 3A: If the 2 sides balance, then the only remaining untested ball is the only possible odd ball. Put it against a normal ball and you have your final answer.
Step 3B: If the 2 sides are not even, then take the 2 balls from the side that did not include a known normal ball. If this side was lighter in step 2 the balls are potentially light, if it was heavier they are potentially heavy. Balance them against each other and whichever one fulfills its potential is the odd ball. If they are even then you know the ball that was with the normal ball is the odd ball. You also know if it is heavy or light based on step 2.
Step 1 possible outcome: The sides are not even. Therefore you now have 4 potentially heavy balls and 4 potentially light balls, and only 2 steps left.
Step 2: Take 2 of the potentially light balls, and 1 of the potentially heavy balls and set them aside. Of the 5 remaining balls put 2 potentially heavy balls and a potentially light ball on 1 side of the balance, and 1 potentially heavy and 1 potentially light on the other. Add 1 normal ball so there are 3 balls on each side.
Step 3A: If the 2 sides are even, then 1 of the 3 balls you set aside must be the odd ball. Put 1 potentially light ball and 1 potentially heavy ball on 1 side of the balance against 2 normal balls. If the side goes down the potentially heavy ball was indeed heavy, if it goes up the potentially light ball was light. If the sides are even then the potentially light ball that is still set aside is indeed light.
Step 3B: If the side with 2 potentially heavy balls and a potentially light ball was heavier in step 2, then you know that either 1 of those 2 potentially heavy balls is the odd ball, or the potentially light ball on the other side of the balance is the odd ball. Therefore you are down to 3 balls, 2 potentially heavy and 1 potentially light, and you can do the same thing you would have done in step 3A.
Step 3C: If the side with 1 potentially heavy ball and 1 potentially light ball was heavier then you know that either the potentially heavy ball is the odd ball, or the potentially light ball on the other side is the odd ball. Therefore you are down to only 2 balls, 1 potentially heavy, 1 potentially light. Put these 2 balls on 1 side of the balance against 2 normal balls and if they are heavier the potentially heavy ball is heavy, and if they are lighter the potentially light ball is light.