2007 AMC 10A Problems/Problem 11

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, 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 $\dfrac{108}{6} = 18\mathrm{(C) }.$

Video Solution

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

~ pi_is_3.14

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 10	Followed by Problem 12
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All AMC 10 Problems and Solutions

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2007 AMC 10A Problems/Problem 11

Contents

Problem

Solution

Solution 2

Video Solution

See also