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}{ | + | 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 16:19, 2 December 2015
Problem
The numbers from 
to 
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?
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 numbers represent all of the vertices of the cube. Thus the answer is 
.
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
