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2007 AMC 10A Problem 11

2007 AMC 10A Problem 11AMC 10 Step by Step

Name: Date: September 13, 2026

Sets, Estimation & Miscellaneous · difficulty 2 of 5 · about 2 min · position P11-15

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?

A)
$14$
B)
$16$
C)
$18$
D)
$20$
E)
$24$

Answer

Key insight

Two opposite faces use every vertex exactly once, so two face sums equal 1 + 2 + ... + 8 = 36 and each face sums to 18.

Solution

Call the common face sum $S$ . Take any face and the face opposite it. Those two faces share no vertices, and between them they use all $8$ vertices of the cube.

So the two face sums together equal the sum of all eight labels:
$2S = 1 + 2 + \cdots + 8 = 36 \quad\Longrightarrow\quad S = 18.$

(Such a labeling exists, for instance top face $1, 4, 5, 8$ in cyclic order with $6, 7, 2, 3$ directly below them, so the question is meaningful.)

The answer is $\boxed{\textbf{(C)}\ 18}$ .

Why this works

When a quantity is constant over several overlapping groups, look for a collection of those groups that covers every element exactly once; then the constant is forced without knowing the arrangement. Opposite faces of a cube are exactly such a partition. Counting the same total two ways is the heart of many "common sum" problems.

Alternative approach

Each vertex lies on $3$ faces, so adding all six face sums counts every label three times: $6S = 3 \cdot 36 = 108$ , hence $S = 18$ .

The trap

Trying to actually construct the labeling instead of noticing that opposite faces partition the vertices.

Common mistakes

Trying to actually construct the labeling instead of noticing that opposite faces partition the vertices.
Dividing $36$ by the number of faces ( $6$ ) or by $4$ , forgetting that each vertex is counted on three faces.

Techniques

Exploit symmetry to reduce work or pair up objects