Difference between revisions of "2007 AMC 10A Problems/Problem 11"
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<math>\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 24</math> | <math>\text{(A)}\ 14 \qquad \text{(B)}\ 16 \qquad \text{(C)}\ 18 \qquad \text{(D)}\ 20 \qquad \text{(E)}\ 24</math> | ||
| − | == Solution == | + | == Solution 1 == |
| − | The sum of the numbers on | + | The sum of the numbers on one face of the cube is equal to the sum of the numbers on the opposite 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 == | ||
| + | Consider a number on a vertex. It will be counted in 3 different faces, since any vertex is the intersection of three edges. Therefore, each number <math>1,2,\cdots,7,8</math> will be added into the total sum <math>3</math> times. Therefore, our total sum is <math>3(1+2+\cdots+8)=108.</math> Finally, since there are <math>6</math> faces, our common sum is <math>\dfrac{108}{6} = 18\mathrm{(C) }.</math> | ||
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| + | == Video Solution by OmegaLearn== | ||
| + | https://youtu.be/ZhAZ1oPe5Ds?t=2075 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== See also == | == See also == | ||
| Line 11: | Line 19: | ||
[[Category:Introductory Combinatorics Problems]] | [[Category:Introductory Combinatorics Problems]] | ||
| + | {{MAA Notice}} | ||
Latest revision as of 19:44, 21 January 2024
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 1
The sum of the numbers on one face of the cube is equal to the sum of the numbers on the opposite 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, since any vertex is the intersection of three edges. 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 by OmegaLearn
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. 
