Difference between revisions of "2014 AMC 10B Problems/Problem 24"
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| − | ==Problem== | + | ==Problem==<math> |
| − | The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is | + | The numbers 1, 2, 3, 4, 5 are to be arranged in a circle. An arrangement is </math>bad<math> if it is not true that for every </math>n<math> from </math>1<math> to </math>15<math> one can find a subset of the numbers that appear consecutively on the circle that sum to </math>n<math>. Arrangements that differ only by a rotation or a reflection are considered the same. How many different bad arrangements are there? |
| − | <math> \textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5 | + | </math> \textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5 $ |
==Solution== | ==Solution== | ||
Revision as of 14:05, 20 February 2014
==Problem==
bad
n
1
15
n
\textbf {(A) } 1 \qquad \textbf {(B) } 2 \qquad \textbf {(C) } 3 \qquad \textbf {(D) } 4 \qquad \textbf {(E) } 5 $
Solution
See Also
| 2014 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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. 
