Fishermen

• Feb 28th 2010, 08:52 PM
harish21
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?
• Feb 28th 2010, 09:56 PM
Soroban
Hello, harish21!

Quote:

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

• Mar 1st 2010, 12:26 AM
harish21
Quote:

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!!
• Mar 1st 2010, 03:08 PM
icemanfan
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.
• Mar 1st 2010, 07:26 PM
Wilmer