# A solution to a fishy affair

Previously, on Minus Two Fish:
P.A.M. Dirac was presented with a puzzle.

Three fishermen come back from the sea. Each collapses in their respective tent. Fisherman #1 wakes up and decides to get his share of the catch. He counts the fish, realizes it is not a number divisible by three, throws away one fish to the sea correcting the situation, and takes a third of the remaining fish into his tent. Fisherman #2 wakes up later and also decides he is going to get his share of the catch. Fisherman #2 wants a third of the fish he sees. It is not a multiple of three, but he throws away one fish and takes a third of the fish and goes to sleep. Fisherman #3 wakes up after, and does the same: he throws away one fish, takes a third of the fish, and goes into his tent.
What is the smallest number of fish for which this would happen?

The real solution was not revealed, until now.

[Dirac’s Fish Puzzle Solution Spoiler Warning]

The want to find what is the smallest number of fish \$\$f\$\$ at the beginning.

\$\$f-1\$\$ is the number of fish after Fisherman #1 throws away one. \$\$left(f-1right)frac{1}{3}\$\$ is the number of fish that Fisherman #1 takes for himself, leaving only \$\$left(f-1right)frac{2}{3}\$\$. \$\$(f-1)frac{2}{3}-1\$\$ is the number of fish after Fisherman #2 throws away one. \$\$left( left(f-1right)frac{2}{3}-1right)frac{2}{3}\$\$ is the number of fish that Fisherman #2 leaves after he takes his share. \$\$left( left( f-1 right) frac{2}{3} -1 right)frac{2}{3}-1 \$\$ is the number of fish after Fisherman #3 throws away one. \$\$left(left(left(f-1right)frac{2}{3}-1right)frac{2}{3}-1right)frac{2}{3}\$\$ is the number of fish that Fisherman #3 leaves after he takes his share.

The last number must be an integer \$\$i\$\$, as it is implicit that only a whole number fish are left.
\$\$i=(((f-1)frac{2}{3}-1)frac{2}{3}-1)frac{2}{3}=ffrac{8}{27}-frac{38}{27}\$\$. Solving for \$\$f\$\$,
\$\$f=ifrac{27}{8}+frac{38}{8}\$\$.

Now, \$\$i\$\$ must be a small integer. The easiest way to explain the solution is by trial and error. So, try \$\$i=1\$\$, and you will see \$\$f\$\$ is not an integer. Try \$\$i=2\$\$, \$\$i=3\$\$ … still \$\$f\$\$ is not an integer. Until you reach \$\$i=6\$\$, where \$\$f=25\$\$.
So, the smallest number of fish \$\$f\$\$ is 25.
Also, try in the equation to plug in \$\$f=-2\$\$, Dirac’s solution and you will see \$\$i=-2\$\$. That is why it is so weird, Dirac assumes that there are negative fish to start, and negative fish at the end.

This solution permits for infinite number of fishermen to come and throw away \$\$1\$\$ fish and take \$\$frac{1}{3}\$\$ of the remaining fish as the number of fish at the end is the same as the beginning.

## Author: minustwofish

I am a quantum physicist.