Let's say it starts wih x fish.
number of fish left after | John | leaves | _ | \(=\ \frac23({\color{Aquamarine}x}-1)\ =\ \frac23x-\frac23\) |
number of fish left after | Joe | leaves |
| \(=\ \frac23(\ {\color{Aquamarine}\frac23x-\frac23}\ -1)\ =\ \frac49x-\frac{10}{9}\) |
number of fish left after | _James_ | leaves |
| \(=\ \frac23(\ {\color{Aquamarine}\frac49x-\frac{10}{9}}\ -1)\ =\ \frac{8}{27}x-\frac{38}{27}\) |
The smallest positive integer value of x that makes \(\frac{8}{27}x-\frac{38}{27}\) a positive integer is 25 .
By looking at a graph we can check this: https://www.desmos.com/calculator/mivvh3t6in
So the minimum possible number of fish before John threw out the first fish is 25 .