Difference between revisions of "2016 AMC 10A Problems/Problem 18"
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==Solution== | ==Solution== | ||
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| + | First of all, the adjacent faces have same sum (18, because 1+2+3+4+5+6+7+8=36, 36/2=18), | ||
| + | consider the the <math>opposite sides</math> (the two sides which are parallel but not in same face of the cube) | ||
| + | they must have same sum value too. | ||
| + | Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of <math>opposite sides</math>, | ||
| + | we should have 1+X=8+Y, but no solution for [2,7], contradiction. | ||
| + | |||
| + | Now we know 1 and 8 must share the same side, which sum is 9, the opposite side also must have sum of 9, same thing for the other two <math>parallel sides</math>. | ||
| + | |||
| + | now we have 4 parallel sides 1-8, 2-7, 3-6, 4-5. | ||
| + | thinking about 4 end points number need to have sum of 18. | ||
| + | it is easy to notice only 1-7-6-4 vs 8-2-3-5 would work. | ||
| + | |||
| + | so if we fix one direction 1-8 (or 8-1) all other 3 parallel sides must lay in one direction. | ||
| + | |||
| + | Now, the problem is same as the problem to arrange 4 points in a 2-D square. which is 4!/4=<math>\boxed{\textbf{(C) }6.}</math> | ||
==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 11:09, 4 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?
Solution
First of all, the adjacent faces have same sum (18, because 1+2+3+4+5+6+7+8=36, 36/2=18),
consider the the 
(the two sides which are parallel but not in same face of the cube)
they must have same sum value too.
Now think about the extreme condition 1 and 8 , if they are not sharing the same side, which means they would become end points of ,
we should have 1+X=8+Y, but no solution for [2,7], contradiction.
Now we know 1 and 8 must share the same side, which sum is 9, the opposite side also must have sum of 9, same thing for the other two 
.
now we have 4 parallel sides 1-8, 2-7, 3-6, 4-5. thinking about 4 end points number need to have sum of 18. it is easy to notice only 1-7-6-4 vs 8-2-3-5 would work.
so if we fix one direction 1-8 (or 8-1) all other 3 parallel sides must lay in one direction.
Now, the problem is same as the problem to arrange 4 points in a 2-D square. which is 4!/4=
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 | ||
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