2023 AMC 10A Problems/Problem 6

Problem

An integer is assigned to each vertex of a cube. The value of an edge is defined to be the sum of the values of the two vertices it touches, and the value of a face is defined to be the sum of the values of the four edges surrounding it. The value of the cube is defined as the sum of the values of its six faces. Suppose the sum of the integers assigned to the vertices is $21$ . What is the value of the cube? $\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$

Solution

Each of the vertices is counted $3$ times because each vertex is shared by three different edges. Each of the edges is counted $2$ times because each edge is shared by two different faces. Since the sum of the integers assigned to all vertices is $21$ , the final answer is $21\times3\times2=\boxed{\textbf{(D) } 126}$

~Mintylemon66

Solution 2

Just set one vertice equal to $21$ , it is trivial to see that there are $3$ faces with value $42$ , and $42 \cdot 3=126$ .

~SirAppel

Solution 3

Since there are 8 verticies in a cube, there are $21/4$ verticies for two edges. There are 4 edges in a face, and 6 faces in a cube, so the value of the cube is $21/4 * 24 = \boxed{\textbf{(D) } 126}$

~DRBStudent ~Failure.Net

Solution 4 (use an example)

Set each vertex to value 1, so the sum of the vertices is 8. We find that the value of the cube, if all vertices are 1, is 48. We conclude that the value of the cube is 6 times the value of the sum of the vertices. Therefore, we choose $21\times6=\boxed{\textbf{(D) } 126}$

~milquetoast

Solution 5

The average value of a vertex on the cube is $\frac{21}{8}$ . Thus, the average value of an edge is $\frac{21}{8} \times2=\frac{21}{4}$ . Since there are 4 edges on each face, the average value of a face is $\frac{21}{4}\times4=\frac{21}{1}=21$ . Since there are 6 faces, our answer is $21\times6=$ $\boxed{\textbf{(D) } 126}$

~Failure.net

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 5	Followed by Problem 7
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All AMC 10 Problems and Solutions

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