AMC 10 · 2023 · #6
Grade 4 geometry-3dProblem
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?
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: A cube has $8$ vertices, $12$ edges, and $6$ faces; Each edge value = sum of its $2$ endpoint vertex values; Each face value = sum of its $4$ surrounding edge values; Sum of all vertex values is $21$; Answer choices: (A) $42$, (B) $63$, (C) $84$, (D) $126$, (E) $252$
Unknowns: The sum of the $6$ face values (the "value of the cube")
Understand
Restated: 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.
Givens: A cube has $8$ vertices, $12$ edges, and $6$ faces; Each edge value = sum of its $2$ endpoint vertex values; Each face value = sum of its $4$ surrounding edge values; Sum of all vertex values is $21$; Answer choices: (A) $42$, (B) $63$, (C) $84$, (D) $126$, (E) $252$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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 $21$.
Execute — Answer: D
K.G.B.4 Step 1 - Sketch a cube and pick one vertex (say the top-front-right corner).
- Count the edges sliding out of it.
- Three edges meet at every vertex — the same is true at all $8$ corners because the cube is symmetric.
💡 Looking at the corner of any box you can see three edges shooting away — a Kindergarten-level observation about 3-D shapes.
4.OA.A.3 Step 2 - Subproblem A: total over all $12$ edges.
- When we sum the $12$ edge values, each vertex value is added once for every edge it sits on — that is $3$ times.
- So the edge-sum is exactly $3 \times (\text{vertex sum})$.
💡 Each corner number gets used by each of its $3$ edges — a Grade 4 multi-step "how many groups" multiplication.
K.G.B.4 Step 3 - Back to the diagram: pick any edge and count the faces that share it.
- Each edge is the boundary between exactly $2$ faces of the cube — true for all $12$ edges.
💡 Run a finger along any edge of a box and you feel the two sides meet there — that's the count, $2$.
4.OA.A.3 Step 4 - Subproblem B: total over all $6$ faces.
- When we sum the $6$ face values, each edge value is added once for every face it borders — that is $2$ times.
- So the face-sum is exactly $2 \times (\text{edge sum})$.
💡 Each edge number gets used by each of its $2$ faces — same "how many groups" multiplication move.
3.OA.C.7 Step 5 - Combine: $S_F = 2 \times 3 \times S_V = 6 \times 21 = 126$.
- That matches choice (D).
💡 Two multiplications in a row stack into one — Grade 3 fluency with multiplication facts finishes the job.
K.G.B.4 Sketch a cube and pick one vertex (say the top-front-right corner). Count the ed 4.OA.A.3 Subproblem A: total over all $12$ edges. When we sum the $12$ edge values, each K.G.B.4 Back to the diagram: pick any edge and count the faces that share it. Each edge 4.OA.A.3 Subproblem B: total over all $6$ faces. When we sum the $6$ face values, each ed 3.OA.C.7 Combine: $S_F = 2 \times 3 \times S_V = 6 \times 21 = 126$. That matches choice Review
Reasonableness: Sanity-check by giving every vertex the value $\tfrac{21}{8}$ (not an integer, but the problem only constrains the total). Each edge then has value $2 \cdot \tfrac{21}{8} = \tfrac{42}{8}$, and the $12$ edges sum to $12 \cdot \tfrac{42}{8} = 63$ — matching $S_E = 63$. Each face's value is the sum of its $4$ edges, each worth $\tfrac{42}{8}$, so a face is $\tfrac{168}{8} = 21$, and $6$ faces sum to $126$. The answer is independent of the particular vertex assignment, as the chain $S_F = 2 \cdot S_E = 2 \cdot 3 \cdot S_V$ shows.
Alternative: Tool #9 (Solve an Easier Related Problem): assign all $8$ vertices the value $1$ and recompute. Then $S_V = 8$, each edge value is $2$, $S_E = 12 \cdot 2 = 24 = 3 S_V$, each face value is $8$, $S_F = 6 \cdot 8 = 48 = 6 S_V$. The ratio $S_F / S_V = 6$ pops out from the small case; scale back: $6 \cdot 21 = 126$.
CCSS standards used (min grade 4)
K.G.B.4Analyze and compare two- and three-dimensional shapes (Reading the cube's structure off a sketch — each vertex touches $3$ edges, each edge touches $2$ faces — the Kindergarten-level observations the whole counting argument rests on.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Turning the two "each vertex is counted $3$ times" and "each edge is counted $2$ times" observations into the multi-step computation $S_E = 3 \cdot S_V$ and $S_F = 2 \cdot S_E$.)3.OA.C.7Fluently multiply and divide within 100 (Doing the final whole-number arithmetic $2 \times 3 \times 21 = 6 \times 21 = 126$.)
⭐ 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 $3 \times 2 = 6$ times the vertex-sum.
⭐ 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 $3 \times 2 = 6$ times the vertex-sum.