Difference between revisions of "2016 AMC 10A Problems/Problem 18"
m (→See Also) |
(→Solution:) |
||
| Line 3: | Line 3: | ||
<math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math> | <math>\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24</math> | ||
| − | |||
==This can be solved by simple counting. I have no idea how to use the editor so I cannot finish this solution.== | ==This can be solved by simple counting. I have no idea how to use the editor so I cannot finish this solution.== | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}} | {{AMC10 box|year=2016|ab=A|num-b=17|num-a=19}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:36, 3 February 2016
Each vertex of a cube is to be labeled with an integer 
through 
, with each integer being used once, in such a way that the sum of the four numbers on the vertices of a face is the same for each face. Arrangements that can be obtained from each other through rotations of the cube are considered to be the same. How many different arrangements are possible?
This can be solved by simple counting. I have no idea how to use the editor so I cannot finish this solution.
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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. 
