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 | + | The sum of the numbers on the top face of the 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>. |
== Solution 2 == | == Solution 2 == |
Revision as of 12:52, 3 June 2021
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 the 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
.
Solution 2
Consider a number on a vertex. It will be counted in 3 different faces, the ones it is on. Therefore, each number will be added into the total sum
times. Therefore, our total sum is
Finally, since there are
faces, our common sum is
Video Solution
https://youtu.be/ZhAZ1oPe5Ds?t=2075
~ pi_is_3.14
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.