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

Latest revision as of 00:44, 10 June 2024

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 1

Let the amount of people be $p$ , horses be $h$ , sheep be $s$ , cows be $c$ , and ducks be $d$ . We know $3p=h$ $4s=c$ $3d=p$ Then the total amount of people, horses, sheep, cows, and ducks may be written as $p+h+s+c+d = 3d+3(3d)+s+4s+d$ . This is equivalent to $13d+5s$ . 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 (Intuitive)

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 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 $13d+5s$ . $13\cdot 5-13-5=47,$ so our answer is $\boxed{\textbf{(B)} 47}$ .

Video Solution

https://youtu.be/SOH98fsk-9c

~savannahsolver

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 5: / Line 5: @@
 <math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math>
-==Solution==
+==Solution 1==
-===Solution 1===
 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>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>.
+We know <cmath>3p=h</cmath> <cmath>4s=c</cmath> <cmath>3d=p</cmath> Then the total amount of people, horses, sheep, cows, and ducks may be written as <math>p+h+s+c+d = 3d+3(3d)+s+4s+d</math>. This is equivalent to <math>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>.
-==Solution 2==
+==Solution 2 (Intuitive)==
-As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as <math>13d+5s</math>. 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 <math>13d+5s</math>.
+As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as <math>13d+5s</math>. However, instead of going through each of the solutions and testing 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 <math>13d+5s</math>.
-<cmath>13*5-13-5=47,</cmath> so our answer is <math>\boxed{\textbf{(B)} 47}</math>.
+<cmath>13\cdot 5-13-5=47,</cmath> so our answer is <math>\boxed{\textbf{(B)} 47}</math>.
+==Video Solution==
+https://youtu.be/SOH98fsk-9c
+~savannahsolver
 ==See Also==
 {{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}}
 {{MAA Notice}}

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