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

Revision as of 11:52, 3 June 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 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 $\dfrac{108}{6}=\mathrm{(C) } 18.$

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
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.

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 == 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 <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>.
+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 ==

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