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2023 AMC 10A Problem 6

2023 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: October 28, 2026

Basic Counting · difficulty 2 of 5 · about 2 min · position P1-10

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?

A) $42$
B) $63$
C) $84$
D) $126$
E) $252$

Answer

Key insight

Follow one vertex's number: 3 edges use it and each edge sits on 2 faces, so it is counted 6 times. The cube's value is 6 times 21.

Solution

We do not know the eight numbers, only their total. So instead of adding things up, follow one vertex's number and count how many times it ends up in the cube's value.

1. Vertex to edges. Each corner of a cube touches $3$ edges. So a vertex's number is added into $3$ edge values.
2. Edges to faces. Each edge of a cube is the border between $2$ faces. So each edge value is added into $2$ face values.
3. Put these together. A vertex's number goes into $3$ edges, and each of those goes into $2$ faces. So the number shows up $3 \times 2 = 6$ times in the sum of the six faces.
4. Every vertex looks the same, so every vertex's number is counted $6$ times. The cube's value is
$6 \times (\text{sum of all vertex numbers}) = 6 \times 21 = 126.$

The answer is $\boxed{\textbf{(D)}\ 126}$ .

Why this works

The value is built in layers: vertices make edges, edges make faces, faces make the cube. Each layer just adds up the one before, so the final total is always the same multiple of the vertex total. To find that multiple, count how many times one vertex is used. Watch for this "how many times is each piece counted" idea whenever a total is built out of shared parts.

Alternative approach

Since only the total $21$ matters, pick easy numbers. Put $21$ on one corner and $0$ on the other seven.

The $3$ edges touching that corner each have value $21$ . All other edges are $0$ .
The $3$ faces touching that corner each contain two of those edges, so each face has value $21 + 21 = 42$ . The other three faces are $0$ .
The cube's value is $3 \times 42 = 126$ .

The trap

Multiplying by 3 for the edges but forgetting that every edge is shared by two faces, which gives 63.

Concise solution

Track how many times a single vertex's number gets counted in the final total.

A vertex belongs to $3$ edges, so its number appears $3$ times in the sum of all edge values. Each edge borders exactly $2$ faces, so when we add up the six face values, every edge is counted twice. Hence each vertex number is counted $3 \cdot 2 = 6$ times in the cube's value.

By symmetry this holds for all eight vertices, so the cube's value is
$6 \cdot (\text{sum of vertex values}) = 6 \cdot 21 = 126.$

The answer is $\boxed{\textbf{(D)}\ 126}$ .

Review in the AoPS books

Introduction to Counting & Probability, Chapter 1: Counting Is Arithmetic
Introduction to Geometry, Chapter 14: Three-Dimensional Geometry
Introduction to Algebra, Chapter 22: Special Manipulations (§22.3)
Introduction to Counting & Probability, Chapter 2: Basic Counting Techniques

Techniques

Test small/specific values or special cases to find or verify the answer · Exploit symmetry to reduce work or pair up objects