Difference between revisions of "2015 AMC 10B Problems/Problem 15"

Revision as of 20:02, 6 March 2015

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

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

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

@@ Line 6: / Line 6: @@
 ==Solution==
+Let the amount of people be <math>p</math>, horses be <math>h</math>, sheep be <math>s</math>, cows be <math>c</math>, and ducks be <math>d</math>.
+We know <cmath>3h=p</cmath> <cmath>4c=s</cmath> <cmath>3p=d</cmath> Then the total amount of people, horses, sheep, cows, and ducks may be written as <math>3h+h+4c+c+9h=13h+5c</math>. Looking through the options, we see <math>47</math> is impossible to make for integer values of <math>h</math> and <math>c</math>. So the answer is <math>\boxed{\textbf{(B)} 47}</math>.
 ==See Also==
 {{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}
 {{MAA Notice}}

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Difference between revisions of "2015 AMC 10B Problems/Problem 15"

Revision as of 20:02, 6 March 2015

Problem

Solution

See Also