Answer to Riddle #65: 9 Coins, 1 Odd one, 2 Weighings

65. There are 9 coins, all except one are the same weight, the odd one is heavier than the rest. You must determine which is the odd one out using an old fashioned balance. You may use the balance twice. Explain how this can be done.

Coins, balls, marbles, old fashioned balance. We've been here before. The key thing here is there are 9 of them, we have 2 weighings and importantly we know the direction of the error, as in the odd one is heavier:

Before reading the answer can I interest you in a clue?

Not nearly as difficult as the the previous two problems of this type. Lets start as always by numbering them 1 to 9.

Put 1,2 & 3 on the left side and 4,5 & 6 on the right side. There are three possible outcomes:
scale tilts left - means the heavy coin is in group 1,2 & 3
scale tilts right - means the heavy coin is in group 4,5 & 6
scale balances - means the heavy coin is in group 7,8 & 9

Using one weighing we have successfully narrowed three groups down to one. If it's not completely clear how we can make the assumption that the bad coins is in group 7,8 & 9 when the scales balance, remember we know there is a bad coin, there has to be, we have eliminated the other two groups, it must be in the third.

We need to do the same again and we will arrive at the final coin. Lets label the three coins we have narrowed it down to as A, B & C. Not strictly necessary but it's easier to explain. So if we'd narrowed it down to 4,5 & 6 they would become A,B & C respectively. Putting A on the left side, B on the right and C to one side.
A goes down - A is the heavy coin
B goes down - B is the heavy coin
Neither goes down - C is the heavy coin

Sometimes this puzzle will be written as the odd coin is lighter than the others. The solution is the basically the same. Narrow it down to the group of three by looking for the side that goes up, then to the specific coin, marble or whatever

I would have anticipated this puzzle would have been right up AI's street. Apparently not.

If you're curious what Bard made of this puzzle...

Sure. Here's how you can do it:

Divide the coins into three groups of three coins each.
Weigh two of the groups against each other.
If the two groups balance, then the odd coin is in the third group.
If the two groups do not balance, then the odd coin is in the heavier group.
Take the heavier group and divide it into two groups of two coins each.
Weigh the two groups against each other.
If the two groups balance, then the odd coin is in the heavier group.
If the two groups do not balance, then the odd coin is in the heavier group.

Here's an example:

1. Divide the coins into three groups of three coins each:
    * A: A1, A2, A3
    * B: B1, B2, B3
    * C: C1, C2, C3
2. Weigh two of the groups against each other:
    * A vs. B
3. If the two groups balance, then the odd coin is in the third group:
    * C
4. If the two groups do not balance, then the odd coin is in the heavier group:
    * A or B
5. Take the heavier group and divide it into two groups of two coins each:
    * A1 vs. A2
6. Weigh the two groups against each other:
    * If the two groups balance, then the odd coin is in the heavier group:
        * A1
    * If the two groups do not balance, then the odd coin is in the heavier group:
        * A2
7. Repeat steps 5-6 until you have identified the odd coin.

If you're curious what ChatGPT made of this puzzle...

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Answer to Riddle #65: 9 Coins, 1 Odd one, 2 Weighings

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