Challenge .....(Long one)

Eargasms wrote:

Ok  Third time lucky then ...........................    Start with 25 throw one away 24 1/3rd taken leaves 16  throw one away 15 1/3rd taken leaves 10  throw one away  9 1/3rd taken away leaves  6  I think i got there in the end ! lol

I assume tha Anothony have got X piece fish, so remain should be 2X.

Below is the formula.

(3X+1) should be the number after Paul got his lot 1/2 *(3X+1), so before Paul got this, the number should be 3/2 *(3X+1)+1.

consequently, total quantity should be 3/2*(3/2 *(3X+1)+1)+1.

we can check the X from 1 to ....., only to ensure that no decimal fraction will appear when calculate this.

Is this right?

i had the same answers as eargasm i just aproached from the other side:

start with the smallest number there is you can divide by three, that being three

That would mean that there would have been 4 when the last fishermen found them, wich would mean there would have been 7 when the middel fishermen found them wich means that the first one would have left 3 eguals from 4 and a half fish, so that one is not possible

then the same with starting number 6, get you to 7 immidiatly so no good either,..and the next one is 9..

If you count that one out you get to 25 to begin with, and with there being 6 left in the end...

So i guess this is also a right 'logical answer'....but is it also the right answer?