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

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

Page

Toolbox

Search

2023 AMC 10A Problems/Problem 6

Contents

Problem

Solution

Solution 2

Solution 3

See Also