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

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== Solution 2 ==
 
== Solution 2 ==
Consider a number on a vertex. It will be counted in 3 different faces, the ones it is on. 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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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>
  
== Video Solution ==
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== Video Solution by OmegaLearn==
 
https://youtu.be/ZhAZ1oPe5Ds?t=2075
 
https://youtu.be/ZhAZ1oPe5Ds?t=2075
  

Revision as of 03:51, 21 January 2023

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

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 $1,2,\cdots,7,8$ will be added into the total sum $3$ times. Therefore, our total sum is $3(1+2+\cdots+8)=108.$ Finally, since there are $6$ faces, our common sum is $\dfrac{108}{6} = 18\mathrm{(C) }.$

Video Solution by OmegaLearn

https://youtu.be/ZhAZ1oPe5Ds?t=2075

~ pi_is_3.14

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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