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2023 AMC 10A Problem 6

2023 AMC 10A Problem 6AMC 10 Step by Step

Name: Date: September 13, 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

Each vertex lies on 3 edges and each edge on 2 faces, so every vertex is counted 6 times: the cube's value is 6 times 21.

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 $\boxed{\textbf{(D)}\ 126}$ .

Why this works

Sums built in layers (vertices to edges to faces to solid) are linear, so the total is a fixed multiple of the vertex sum, and that multiple is just the number of times one vertex gets counted. Counting incidences (how many edges per vertex, how many faces per edge) replaces any need to know the individual numbers.

Alternative approach

Put $21$ on one vertex and $0$ on the other seven. The three edges at that vertex each have value $21$ ; each of those edges lies on two faces, so the three faces meeting at that vertex each have value $42$ , and the other faces are $0$ . Total $3 \cdot 42 = 126$ .

The trap

Multiplying by 3 (edges per vertex) but forgetting that each edge is also counted by two faces, giving 63.

Common mistakes

Multiplying by 3 (edges per vertex) but forgetting that each edge is also counted by two faces, giving 63.
Multiplying by $12$ edges or $6$ faces directly, as if every edge contained every vertex.

Techniques

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