AMC 10 · 2023 · #6

Easy mode Grade 4

Problem

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?

$\textbf{(A) } 42 \qquad \textbf{(B) } 63 \qquad \textbf{(C) } 84 \qquad \textbf{(D) } 126 \qquad \textbf{(E) } 252$

Pick an answer.

(A)

(B)

(C)

(D)

126

(E)

252

View mode:

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

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

#1 Draw a Diagram 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.

$$\text{edges meeting at each vertex} = 3$$

💡 Looking at the corner of any box you can see three edges shooting away — a Kindergarten-level observation about 3-D shapes.

#7 Identify Subproblems 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 $3 \times (\text{vertex sum})$.

$$S_E = 3 \times S_V = 3 \times 21 = 63$$

💡 Each corner number gets used by each of its $3$ edges — a Grade 4 multi-step "how many groups" multiplication.

#1 Draw a Diagram 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.

$$\text{faces sharing each edge} = 2$$

💡 Run a finger along any edge of a box and you feel the two sides meet there — that's the count, $2$.

#7 Identify Subproblems 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 $2 \times (\text{edge sum})$.

$$S_F = 2 \times S_E = 2 \times 63 = 126$$

💡 Each edge number gets used by each of its $2$ faces — same "how many groups" multiplication move.

#7 Identify Subproblems 3.OA.C.7 Step 5

Combine: $S_F = 2 \times 3 \times S_V = 6 \times 21 = 126$.
That matches choice (D).

$$S_F = 6 \times 21 = 126 \;\Rightarrow\; \textbf{(D)}$$

💡 Two multiplications in a row stack into one — Grade 3 fluency with multiplication facts finishes the job.

[1] #1 K.G.B.4 Sketch a cube and pick one vertex (say the top-front-right corner). Count the ed

[2] #7 4.OA.A.3 Subproblem A: total over all $12$ edges. When we sum the $12$ edge values, each

[3] #1 K.G.B.4 Back to the diagram: pick any edge and count the faces that share it. Each edge

[4] #7 4.OA.A.3 Subproblem B: total over all $6$ faces. When we sum the $6$ face values, each ed

[5] #7 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.4 Analyze 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.3 Solve 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.7 Fluently 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.