## Recreational Math

You have 120 coins, all of which are visually identical. E
Except for one coin, they all weigh exactly the same.
It is unknown whether the odd weight coin is lighter or heavier then the others.
Using a two pan balance just five times, identify the odd weight coin.
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 This is one the answer: Spoiler 120 / 3 = 40 ---> 40a, 40b, 40c Balance 1 40a ---- 40b Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier 40 / 3 = 15a, 15b, 10 Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3 10 / 3 = 5a, 5b Balance 4 5a ---- 5b Balance 5 2a ---- 2b

## Recreational Math

 Quote by LAF This is one the answer: Spoiler 120 / 3 = 40 ---> 40a, 40b, 40c Balance 1 40a ---- 40b Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier 40 / 3 = 15a, 15b, 10 Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3 10 / 3 = 5a, 5b Balance 4 5a ---- 5b Balance 5 2a ---- 2b
This solution reduces it to 2 or 3 coins. Neither identifies the one coin.

 Quote by LAF This is one the answer: Spoiler 120 / 3 = 40 ---> 40a, 40b, 40c Balance 1 40a ---- 40b Balance 2 40b ---- 40c = you know the 40s different and if they are lighter or heavier 40 / 3 = 15a, 15b, 10 Balance 3 15a ---- 15b = if equal, then Balance 4 is 7----7 and Balance 5 is 3---3 10 / 3 = 5a, 5b Balance 4 5a ---- 5b Balance 5 2a ---- 2b
Spoiler

Your second balance has the chance of being redundant. If group A and B weigh the same, the coin is in group C.

 Quote by flatmaster Spoiler Your second balance has the chance of being redundant. If group A and B weigh the same, the coin is in group C.
Well a solution has to assume the worst possible scenario

 Quote by jvwert You have 120 coins, all of which are visually identical. E Except for one coin, they all weigh exactly the same. It is unknown whether the odd weight coin is lighter or heavier then the others. Using a two pan balance just five times, identify the odd weight coin.
In fact, it can be done for 121 coins.
 Please remove non-identifies coin which is 2 or 3. Then, you get answered.
 Using your solution to the similar 12 coin problem as a template jvwert: Spoiler First split the 120 coins into main groups of 40, and split these into subgroups of 9, 27, 3, and 1 coins. First weighing: Place one main group on each pan of the balance and leave the other on the table. Second weighing: Rotate the 27 coin sub groups (from pan A to the table, pan B to pan A, and from the table to pan B) and observe whether the balance changes. If it does it will identify a 27 coin subgroup that contains the odd coin and whether it is heavy or light. Clear the balance and split this subgroup into three groups of nine, and for the third weighing place one of these groups of nine on each pan of the balance. This will identify the group of nine that contains the odd coin. Split this group into three groups of three. The solution is then trivial. If the balance does not change then for the third weighing rotate the 9 coin subgroups and observe the balance. If it changes this will identify a 9 coin subgroup that contains the odd coin and whether it is light or heavy. Clear the balance and split this group into three groups of three. The solution is then trivial. If it does not change then for the fourth weighing rotate the 3 coin subgroups and observe the balance. If it changes this will identify a 3 coin subgroup that contains the odd coin and reveal its relative weight. Clear the balance and split this group into ones. The solution is then trivial. If it does not change then for the fifth weighing rotate the 1 coin subgroups which will identify the odd coin and its relative weight.