AMC 10 · 2023 · #6
Easy mode Grade 4Problem
Picture a cube. It has 8 corners, 12 edges, and 6 flat faces.
Someone writes a whole number at each of the 8 corners. Then they make up two scoring rules.
The score of an edge is the sum of the two corner numbers at its ends. The score of a face is the sum of the scores of the 4 edges around it. The score of the cube is the sum of the scores of all 6 faces.
Suppose the 8 corner numbers add up to 21. What is the score 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.