2015 AMC 10B Problems/Problem 15

Revision as of 19:02, 6 March 2015 by Tkhalid (talk | contribs) (Solution)

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

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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png