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

m (Solution)
m (Solution)
Line 7: Line 7:
 
==Solution==
 
==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>.  
 
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>h=3p</cmath> <cmath>c=4s</cmath> <cmath>p=3d</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>.
+
We know <cmath>h=3p</cmath> <cmath>c=4s</cmath> <cmath>p=3d</cmath> Then the total amount of people, horses, sheep, cows, and ducks may be written as <math>3p+p+4s+s+9d=13d+5s</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==
 
==See Also==
 
{{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}
 
{{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:24, 25 December 2017

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

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