Difference between revisions of "2015 AMC 10B Problems/Problem 15"
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==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> | + | 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>. |
==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 12:23, 25 December 2017
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
Let the amount of people be , horses be , sheep be , cows be , and ducks be . We know Then the total amount of people, horses, sheep, cows, and ducks may be written as . Looking through the options, we see is impossible to make for integer values of and . So the answer is .
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 |
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