Answer:
The minimum possible initial amount of fish:![52](https://tex.z-dn.net/?f=52)
Step-by-step explanation:
Let's start by saying that
= is the initial number of fishes
John:
When John arrives:
- he throws away one fish from the bunch
![x-1](https://tex.z-dn.net/?f=x-1%20)
- divides the remaining fish into three.
![\dfrac{x-1}{3} + \dfrac{x-1}{3} + \dfrac{x-1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx-1%7D%7B3%7D%20%2B%20%5Cdfrac%7Bx-1%7D%7B3%7D%20%2B%20%5Cdfrac%7Bx-1%7D%7B3%7D)
- takes a third for himself.
![\dfrac{x-1}{3} + \dfrac{x-1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7Bx-1%7D%7B3%7D%20%2B%20%5Cdfrac%7Bx-1%7D%7B3%7D)
the remaining fish are expressed by the above expression. Let's call it John
![\text{John}=\dfrac{x-1}{3} + \dfrac{x-1}{3}](https://tex.z-dn.net/?f=%5Ctext%7BJohn%7D%3D%5Cdfrac%7Bx-1%7D%7B3%7D%20%2B%20%5Cdfrac%7Bx-1%7D%7B3%7D)
and simplify it!
![\text{John}=\dfrac{2x}{3} - \dfrac{2}{3}](https://tex.z-dn.net/?f=%5Ctext%7BJohn%7D%3D%5Cdfrac%7B2x%7D%7B3%7D%20-%20%5Cdfrac%7B2%7D%7B3%7D)
When Joe arrives:
- he throws away one fish from the remaining bunch
![\text{John} -1](https://tex.z-dn.net/?f=%5Ctext%7BJohn%7D%20-1)
- divides the remaining fish into three
![\dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3} + \dfrac{\text{John} -1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D%20%2B%20%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D%20%2B%20%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D)
- takes a third for himself.
![\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}](https://tex.z-dn.net/?f=%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D%2B%20%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D)
the remaining fish are expressed by the above expression. Let's call it Joe
![\text{Joe}=\dfrac{\text{John} -1}{3}+ \dfrac{\text{John} -1}{3}](https://tex.z-dn.net/?f=%5Ctext%7BJoe%7D%3D%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D%2B%20%5Cdfrac%7B%5Ctext%7BJohn%7D%20-1%7D%7B3%7D)
and simiplify it
![\text{Joe}=\dfrac{2}{3}(\text{John}-1)](https://tex.z-dn.net/?f=%5Ctext%7BJoe%7D%3D%5Cdfrac%7B2%7D%7B3%7D%28%5Ctext%7BJohn%7D-1%29)
since we've already expressed John in terms of x, we express the above expression in terms of x as well.
![\text{Joe}=\dfrac{2}{3}\left(\dfrac{2x}{3} - \dfrac{2}{3}-1\right)](https://tex.z-dn.net/?f=%5Ctext%7BJoe%7D%3D%5Cdfrac%7B2%7D%7B3%7D%5Cleft%28%5Cdfrac%7B2x%7D%7B3%7D%20-%20%5Cdfrac%7B2%7D%7B3%7D-1%5Cright%29)
![\text{Joe}=\dfrac{4x}{9} - \dfrac{10}{9}](https://tex.z-dn.net/?f=%5Ctext%7BJoe%7D%3D%5Cdfrac%7B4x%7D%7B9%7D%20-%20%5Cdfrac%7B10%7D%7B9%7D)
When James arrives:
We're gonna do this one quickly, since its the same process all over again
![\text{James}=\dfrac{\text{Joe} -1}{3}+ \dfrac{\text{Joe} -1}{3}](https://tex.z-dn.net/?f=%5Ctext%7BJames%7D%3D%5Cdfrac%7B%5Ctext%7BJoe%7D%20-1%7D%7B3%7D%2B%20%5Cdfrac%7B%5Ctext%7BJoe%7D%20-1%7D%7B3%7D)
![\text{James}=\dfrac{2}{3}\left(\dfrac{4x}{9} - \dfrac{10}{9}-1\right)](https://tex.z-dn.net/?f=%5Ctext%7BJames%7D%3D%5Cdfrac%7B2%7D%7B3%7D%5Cleft%28%5Cdfrac%7B4x%7D%7B9%7D%20-%20%5Cdfrac%7B10%7D%7B9%7D-1%5Cright%29)
![\text{James}=\dfrac{8x}{27} - \dfrac{38}{27}](https://tex.z-dn.net/?f=%5Ctext%7BJames%7D%3D%5Cdfrac%7B8x%7D%7B27%7D%20-%20%5Cdfrac%7B38%7D%7B27%7D)
This is the last remaining pile of fish.
We know that no fish was divided, so the remaining number cannot be a decimal number. <u>We also know that this last pile was a multiple of 3 before a third was taken away by James</u>.
Whatever the last remaining pile was (let's say
), a third is taken away by James. the remaining bunch would be ![\frac{n}{3}+\frac{n}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bn%7D%7B3%7D%2B%5Cfrac%7Bn%7D%7B3%7D)
hence we've expressed the last pile in terms of n as well. Since the above 'James' equation and this 'n' equation represent the same thing, we can equate them:
![\dfrac{n}{3}+\dfrac{n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}](https://tex.z-dn.net/?f=%5Cdfrac%7Bn%7D%7B3%7D%2B%5Cdfrac%7Bn%7D%7B3%7D%3D%5Cdfrac%7B8x%7D%7B27%7D%20-%20%5Cdfrac%7B38%7D%7B27%7D)
![\dfrac{2n}{3}=\dfrac{8x}{27} - \dfrac{38}{27}](https://tex.z-dn.net/?f=%5Cdfrac%7B2n%7D%7B3%7D%3D%5Cdfrac%7B8x%7D%7B27%7D%20-%20%5Cdfrac%7B38%7D%7B27%7D)
L.H.S must be a Whole Number value and this can be found through trial and error. (Just check at which value of n does 2n/3 give a non-decimal value) (We've also established from before that n is a multiple a of 3, so only use values that are in the table of 3, e.g 3,6,9,12,..
at n = 21, we'll see that 2n/3 is a whole number = 14. (and since this is the value of n to give a whole number answer of 2n/3 we can safely say this is the least possible amount remaining in the pile)
![14=\dfrac{8x}{27} - \dfrac{38}{27}](https://tex.z-dn.net/?f=14%3D%5Cdfrac%7B8x%7D%7B27%7D%20-%20%5Cdfrac%7B38%7D%7B27%7D)
by solving this equation we'll have the value of x, which as we established at the start is the number of initial amount of fish!
![14=\dfrac{8x}{27} - \dfrac{38}{27}](https://tex.z-dn.net/?f=14%3D%5Cdfrac%7B8x%7D%7B27%7D%20-%20%5Cdfrac%7B38%7D%7B27%7D)
![x=52](https://tex.z-dn.net/?f=x%3D52)
This is minimum possible amount of fish before John threw out the first fish