# 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, 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

~ pi_is_3.14

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