AMC 10 · 2023 · #6

Grade 4 geometry-3d
spatial-visualizationface-adjacencymulti-digit-arithmetic identify-subproblemsdouble-counting ↑ Prerequisites: spatial-visualizationmulti-digit-arithmetic
📏 Medium solution 💡 2 insights
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Problem
Each of the 8 vertices of a cube gets a number, and those numbers sum to 21. The cube has 12 edges (each scoring the sum of its two endpoints) and 6 faces (each scoring the sum of its four surrounding edges). Find the sum of all six face values.

Pick an answer.

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

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Draw a Diagram

Tool #1 (Draw a Diagram) is the lead because the trigger is "cube" — sketching it lets you see (and count) how many edges meet at a vertex and how many faces share an edge. Tool #7 (Identify Subproblems) cleanly splits the chain into two count-the-overlap subproblems: first "how many edges does each vertex sit on?" then "how many faces does each edge sit on?" Each subproblem is one whole-number multiplication, and the final answer is the product of those two multipliers times 21.

1STEP 1

Sketch the cube: at every corner exactly 3 edges meet — same at all 8 vertices.

edges meeting at each vertex = 3
2STEP 2

Summing all 12 edges counts each vertex value 3 times, so the edge total is 3 × 21 = 63.

S_E = 3 × S_V = 3 × 21 = 63
3STEP 3

Now trace any edge: exactly 2 faces meet along it — true for all 12 edges.

faces sharing each edge = 2
4STEP 4

Summing all 6 faces counts each edge value 2 times, so the face total is 2 × 63 = 126.

S_F = 2 × S_E = 2 × 63 = 126
5STEP 5

Chaining both: the face total is 2 × 3 × 21 = 6 × 21 = 126, matching choice (D).

S_F = 6 × 21 = 126 → (D)
Answer
126
Sanity-check by giving every vertex the value 218\frac{21}{8} (not an integer, but the problem only constrains the total). Each edge then has value 2 · 218\frac{21}{8} = 428\frac{42}{8}, and the 12 edges sum to 12 · 428\frac{42}{8} = 63 — matching S_E = 63. Each face's value is the sum of its 4 edges, each worth 428\frac{42}{8}, so a face is 1688\frac{168}{8} = 21, and 6 faces sum to 126. The answer is independent of the particular vertex assignment, as the chain S_F = 2 · S_E = 2 · 3 · S_V shows.
💡Key takeaway

This AMC 10 problem only needs the Grade 4 multi-step multiplication idea you already know — each corner number gets passed up to 3 edges, each edge number gets passed up to 2 faces, so the cube's face-sum is 3 × 2 = 6 times the vertex-sum.