1. ## Fishermen

Three fisherman come back from a day of fishing and put their fish together and go to bed. The 1st one wakes up goes to the pile of fish, throws one away and takes a 3rd and leaves. The 2nd one does the same. The last one does the same as well. What is the minimum number of fish that this can work with?

2. Hello, harish21!

Three fisherman come back from a day of fishing and put their fish together and go to bed.
The 1st one wakes up, goes to the pile of fish, throws one away and takes a 3rd and leaves.
The 2nd one does the same. The last one does the same as well.
What is the minimum number of fish that this can work with?
Let $N$ = total number of fish.

The 1st discards one fish and take a third of the remainder.
Then $N$ must be of the form: . $3A + 1$
. . $N \:=\:3A + 1$ .[1]

He discards one fish . . . $3A$ fish left.
He take a third of the fish ( $A$ fish) and leaves.
There are $2A$ fish left.

The 2nd discards one fish and takes a third of the remainder.
Then $2A$ must be of the form $3B + 1$
. . $2A \:=\:3B+1$ .[2]

He discards one fish . . . $3B$ fish left.
He take third of the fish ( $B$ fish) and leaves.
There are $2B$ fish left.

The 3rd discards one fish and takes a third of the remainder.
Then $2B$ must be of the form $3C + 1$
. . $2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2}$ .[3]

Substitute [3] into [2]: . $2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4}$ .[4]

Substitute [4] into [1]: . $N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}$

We have: . $N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}$

Hence: . $N \;=\;6C+4 + \frac{3(C+1)}{4}$

Since $N$ is an integer, $C+1$ must be divisible by 4.
The first time this happens is when $C = 3.$

Therefore: . $N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}$

3. Originally Posted by Soroban
Hello, harish21!

Let $N$ = total number of fish.

The 1st discards one fish and take a third of the remainder.
Then $N$ must be of the form: . $3A + 1$
. . $N \:=\:3A + 1$ .[1]

He discards one fish . . . $3A$ fish left.
He take a third of the fish ( $A$ fish) and leaves.
There are $2A$ fish left.

The 2nd discards one fish and takes a third of the remainder.
Then $2A$ must be of the form $3B + 1$
. . $2A \:=\:3B+1$ .[2]

He discards one fish . . . $3B$ fish left.
He take third of the fish ( $B$ fish) and leaves.
There are $2B$ fish left.

The 3rd discards one fish and takes a third of the remainder.
Then $2B$ must be of the form $3C + 1$
. . $2B \:=\:3C + 1 \quad\Rightarrow\quad B \:=\:\frac{3C+1}{2}$ .[3]

Substitute [3] into [2]: . $2A \;=\;3\left(\frac{3C+1}{2}\right) + 1 \quad\Rightarrow\quad A \:=\:\frac{9C+5}{4}$ .[4]

Substitute [4] into [1]: . $N \;=\;3\left(\frac{9C+15}{4}\right) + 1 \;=\;\frac{27C + 19}{4}$

We have: . $N \;=\;\frac{24C + 16 + 3C + 3}{4} \;=\;\frac{24C+16}{4} + \frac{3C+3}{4}$

Hence: . $N \;=\;6C+4 + \frac{3(C+1)}{4}$

Since $N$ is an integer, $C+1$ must be divisible by 4.
The first time this happens is when $C = 3.$

Therefore: . $N \;=\;6(3) + 4 + \frac{3(3+1)}{4} \quad\Rightarrow\quad \boxed{N \:=\:25}$

Awesome Soroban! I was given this problem in one of my classes and found it very interesting. I thought it would be nice to share. 25 is indeed the logical answer. However -2 is also a correct answer to this problem. I forgot the name of the scientist who gave an answer of -2 to this problem!!

4. If the total number of fish is 25, then:

The first person throws one away (total down to 24) and takes a third (total down to 16).

The second person throws one away (total down to 15) and takes a third (total down to 10).

The third person throws one away (total down to 9) and takes a third (total down to 6).

If every time someone takes a third of the fish, the number remaining must be a whole number, then 25 is the minimum number. Without that restriction, the minimum number of fish is 4.75. In both cases I assume you're not allowed to have a negative number of fish.