2016 AMC 10A Problems/Problem 18

Each vertex of a cube is to be labeled with an integer $1$ through $8$ , 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?

$\textbf{(A) } 1\qquad\textbf{(B) } 3\qquad\textbf{(C) }6 \qquad\textbf{(D) }12 \qquad\textbf{(E) }24$

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 $opposite sides$ (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 $opposite sides$ , 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 parallel sides.

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= $\boxed{\textbf{(C) }6.}$

2016 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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2016 AMC 10A Problems/Problem 18

Solution

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