Let $a$ be the number of fish. Then you know that $a=3b+1$ for some non-negative integer $b$, since Ed saw that the remainder was one when he tried to divide the fish into thirds. He threw one back and took $b$ or himself, leaving $2b$. Similarly, $2b=3c+1$ and $2c=3d+1$, where $d$ is the number of fish Eric ended up taking at the end.
We know that $d$ must be odd, so equal to $2k+1$ for some non-negative $k$. Plugging that in we get
$$2c=6k+4$$
or
$$c=3k+2.$$
Therefore
$$2b = 9k+7$$
and $k$ must also be odd, $k=2l+1$. Then
$$2b = 18l+16$$
or
$$b = 9l+8$$
and
$$a = 27l +25.$$
The smallest value of $a$ occurs when $l=0$: Ed sees 25 fish, tosses one back, and takes 8 (leaving 16). Eddie tosses another back, taking five and leaving 10. Lastly Eric tosses one and takes three.
By using mixed integer linear programming, I obtained an optimal solution with a total profit of $618.56, achieved by the following schedule, where time is measured in hours starting from 0 and ending at 8:
town rate dropoff pickup duration profit
Ellistown 22 0.5099 7.4901 6.9802 153.56
Brownvale 26 0.80722 7.1928 6.3856 166.03
Aliceville 22 1.2737 6.7263 5.4526 119.96
Carlstown 21 1.85404 6.146 4.2919 90.13
Greenvale 27 2.35404 5.646 3.2919 88.88
The formulation uses two nodes $(i,1)$ and $(i,2)$ for each town $i$, for dropoff and pickup, respectively. Let $p_i$ denote the hourly profit rate for town $i$, and let $t_{i,j}$ denote the travel time (in hours) from town $i$ to town $j$. Let binary decision variable $x_i$ indicate whether town $i$ is visited. Let binary decision variable $y_{i,k_i,j,k_j}$ indicate whether node $(i,k_i)$ is visited immediately before node $(j,k_j)$. Let $\overline{a}_i = t_{i,\text{Hometown}}$, and let decision variable $a_{i,k} \in [0,\overline{a}_i]$ be the arrival time at node $(i,k)$. The problem is to minimize $$\sum_i p_i (a_{i,2}-a_{i,1})$$ subject to
\begin{align}
x_{\text{Hometown}} &= 1 \\
\sum_i x_{i} &= 6 \\
\sum_{j,k_j} y_{i,k_i,j,k_j} &= x_{i} &&\text{for $i,k_i$} \\
\sum_{j,k_j} y_{j,k_j,i,k_i} &= x_{i} &&\text{for $i,k_i$} \\
\sum_{i,j} y_{i,1,j,2} &= 1 \\
a_{i,k} &\le \overline{a}_i x_i &&\text{for $i,k$}\\
a_{i,k_i} + t_{i,j} - a_{j,k_j} &\le (\overline{a}_i + t_{i,j}) (1 - y_{i,k_i,j,k_j}) && \text{for $i,k_i,j,k_j$ with $j \not=$ Hometown}
\end{align}
Constraints 3 and 4 are flow balance constraints. Constraint 5 forces exactly one transition from dropoff to pickup, which happens at Greenvale in the optimal solution. The final "MTZ" constraints prevent subtours and enforce $a_{i,k_i} + t_{i,j} \le a_{j,k_j}$ if $y_{i,k_i,j,k_j}=1$.
Best Answer
If $u>v>w$ are the number of chickens that the three farmers sold at the higher price, then, by linearity*, $(26-16)(u-v) = (16-10)(v-w)$, and so $u-v$ is divisible by $3$, and $v-w$ is divisible by $5$. It follows that $(u,v,w) \in \{(8,5,0),(9,6,1), (10,7,2)\}$.
Only the middle solution gives prices that can be measured in USD (or most currencies), namely $\$3.75$ and $\$1.25$.
* To elaborate on this point: If a farmer begins with $n$ chickens, sells $f(n)$ at the higher price point $A$ and $n-f(n)$ at the lower price point $B$, and earns a profit $P$, then the points $(n,f(n))$ are collinear (assuming that $A,B,P$ are constant).
This is because these points are the solutions to a linear equation in two variables, namely $Af(n) + B(n-f(n)) = P$. A line has constant slope, so the equation in my first paragraph follows.