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> | ||
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| + | ==Solution== | ||
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| + | ==See Also== | ||
| + | {{AMC10 box|year=2015|ab=B|num-b=14|num-a=16}} | ||
| + | {{MAA Notice}} | ||
Revision as of 12:01, 5 March 2015
Problem 15
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
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. 
