To start with, I had no idea how to solve this problem. I asked Vincent a bunch of questions and still had really no idea of how to solve it. After talking to Ainsley, she gave me the hint of looking at it from a computer science context. From there, I was able to come up with an answer! Although, this question is really interesting because it seems to have many ways to solve it, which would be really interesting in a classroom context to develop logical thinking.
Once I started thinking of looking at the combination of rats that get sick to figure out which bottle is poisoned, it all made sense!
With 2 rats, you can check 4 bottles. If no rats get sick, it’s bottle 1, if rat A gets sick, it’s bottle 2, if rat B gets sick, it’s bottle 3 and if both get sick, it’s bottle 4.
Extending this to 3 rats, I found
Rat A. Rat B. Rat C
1. 0. 0. 0
2. 1. 0. 0
3. 0. 1. 0
4. 0. 0. 1
5. 1. 1. 0
6. 1. 0. 1
7. 0. 1. 1
8. 1. 1. 1
These are all the combinations with 3 rats, therefore there are 8 bottles that we can check with 3 rats. It seems to be going up by powers of 2, since 2^3=8
2^10=1024, so we would be able to tell 1000 bottles by using the combination of rats that got sick to identify the poisoned bottle.
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