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

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 $\boxed{\textbf{(B)} 47}$ .

2015 AMC 10B (Problems • Answer Key • Resources)
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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2015 AMC 10B Problems/Problem 15

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