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

(→‎Solution: I changed the 4 in the denominator into a 2, so the solution would actually make sense.)
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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}=\mathrm{(C) } 18.</math>
 
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}=\mathrm{(C) } 18.</math>
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== Video Solution ==
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https://youtu.be/ZhAZ1oPe5Ds?t=2075
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~ pi_is_3.14
  
 
== See also ==
 
== See also ==

Revision as of 21:18, 17 January 2021

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 $\frac{1 + 2 + \cdots + 8}{2} = 18\ \mathrm{(C)}$.

Solution 2

Consider a number on a vertex. It will be counted in 3 different faces, the ones it is on. 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 $\dfrac{108}{6}=\mathrm{(C) } 18.$


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

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