2015 AMC 10B Problem 15

If there are $h$ horses and $c$ cows, then there are $3h$ people, $9h$ ducks, and $4c$ sheep. The total is therefore $13h+5c$. The listed values except $47$ can be written in that form: $$\begin{gathered} 41=13\cdot2+5\cdot3, \\ 59=13\cdot3+5\cdot4, \\ 61=13\cdot2+5\cdot7, \\ 66=13\cdot2+5\cdot8. \end{gathered}$$ For $47$, subtracting $0,13,26,39$ leaves $47,34,21,8$, none of which is divisible by $5$. Thus, the correct answer is B.