2015 AMC 10B Problems/Problem 15

Problem

The town of Hamlet has $3$ people for each horse, $4$ sheep for each cow, and $3$ ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?

$\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66$

Solution

Solution 1

Let the amount of people be $p$, horses be $h$, sheep be $s$, cows be $c$, and ducks be $d$. We know \[h=3p\] \[c=4s\] \[p=3d\] Then the total amount of people, horses, sheep, cows, and ducks may be written as $3p+p+4s+s+9d=13d+5s$. Looking through the options, we see $47$ is impossible to make for integer values of $h$ and $c$. So the answer is $\boxed{\textbf{(B)} 47}$.

Solution 2

As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as $13d+5s$. However, instead of going through each of the solutions and testing the the options, you can use the Chicken McNugget Theorem to find the greatest number of people, horses, sheep, cows, and ducks that cannot be written in the form $13d+5s$. \[13*5-13-5=47,\] so our answer is $\boxed{\textbf{(B)} 47}$.

See Also

2015 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Problem 16
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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