Difference between revisions of "2015 AMC 10B Problems/Problem 15"
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<math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math> | <math>\textbf{(A) }41\qquad\textbf{(B) }47\qquad\textbf{(C) }59\qquad\textbf{(D) }61\qquad\textbf{(E) }66</math> | ||
| − | ==Solution== | + | ==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>. | 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 | + | 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 (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 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\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== | ==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}} | ||
Latest revision as of 00:44, 10 June 2024
Problem
The town of Hamlet has 
people for each horse, 
sheep for each cow, and ducks for each person. Which of the following could not possibly be the total number of people, horses, sheep, cows, and ducks in Hamlet?
Solution 1
Let the amount of people be 
, horses be 
, sheep be 
, cows be 
, and ducks be 
.
We know ![\[3p=h\]](http://latex.artofproblemsolving.com/8/e/c/8ec9caac505df773ceb3647f5e311ba8f92e07cb.png)
![\[4s=c\]](http://latex.artofproblemsolving.com/4/8/1/481f33f8826a3655d4e42927cae977027388206b.png)
![\[3d=p\]](http://latex.artofproblemsolving.com/d/4/3/d43f3ef835d562db6c6473375929a629c00732ab.png)
Then the total amount of people, horses, sheep, cows, and ducks may be written as 
. This is equivalent to 
. Looking through the options, we see 
is impossible to make for integer values of and . So the answer is 
.
Solution 2 (Intuitive)
As the solution above says, the total amount of people, horses, sheep, cows, and ducks may be written as . 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 .
![\[13\cdot 5-13-5=47,\]](http://latex.artofproblemsolving.com/e/3/c/e3cc9ce8306192c97c0dee56b766eabe800d4efb.png)
so our answer is .
Video Solution
~savannahsolver
See Also
| 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. 
