Difference between revisions of "2007 AMC 10A Problems/Problem 11"

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== Solution ==
 
== Solution ==
The sum of the numbers on the top face of a cube is equal to the sum of the numbers on the bottom face of the cube; these <math>8</math> numbers represent all of the vertices of the cube. Thus the answer is <math>\frac{1 + 2 + \cdots + 8}{2} = 18\ \mathrm{(C)}</math>.
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The sum of the numbers on the top face of a cube is equal to the sum of the numbers on the bottom face of the cube; these <math>8</math> numbers represent all of the vertices of the cube. Thus the answer is <math>\frac{1 + 2 + \cdots + 8}{4} = 18\ \mathrm{(C)}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 17:19, 2 December 2015

Problem

The numbers from $1$ to $8$ are placed at the vertices of a cube in such a manner that the sum of the four numbers on each face is the same. What is this common sum?

$\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 24$

Solution

The sum of the numbers on the top face of a cube is equal to the sum of the numbers on the bottom face of the cube; these $8$ numbers represent all of the vertices of the cube. Thus the answer is $\frac{1 + 2 + \cdots + 8}{4} = 18\ \mathrm{(C)}$.

See also

2007 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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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