AMC 8 · 2025 · #21

Grade 6 logiccounting

Archetype Constraint Graph Coloring └ Sub-archetype Degree-Driven Extreme Pinning

logical-deductionsystematic-enumerationinterval-arithmetic caseworklogical-deductionsystematic-enumeration ↑ Prerequisites: logical-deductionsystematic-enumeration

📏 Long solution 💡 4 insights 📊 Diagram

Problem

Seven pods A through G need to be assigned the seven grade levels 1 through 7, one grade per pod. Some pods are joined by walkways, and any two pods joined directly must receive grades that differ by at least 2. Find the sum g(C) + g(E) + g(F).

Pick an answer.

(A)

~12

(B)

~13

(C)

~14

(D)

~15

(E)

~16

AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Eliminate Possibilities

The graph already gives a picture, so Tool #1 just means "read the diagram carefully and list every walkway." Tool #15 reorganizes that walkway list into a degree count per pod, which exposes that pods C and F are the most constrained (degree 5). Tool #3 then drives the whole solution: for the highly constrained pods, most candidate grades are eliminated by a simple counting argument (a pod with 5 neighbors needs 5 usable grades among the other 6, and only grades 1 and 7 leave that many), and the same elimination logic then forces g(D), g(G), and finally g(E) one at a time.

1STEP 1

Read the figure and list every connection — there are 12 walkways in all.

12 walkways total

2STEP 2

Recount the walkways as a degree per pod; only C and F reach degree 5, so they are the most constrained.

deg(A)=4, deg(B)=3, deg(C)=5, deg(D)=2, deg(E)=3, deg(F)=5, deg(G)=2

3STEP 3

A degree-5 pod with grade k needs 5 usable grades for its neighbors; only k = 1 or k = 7 leaves that many.

k=1 :{3,4,5,6,7} (5 ✓) | k=2 :{4,5,6,7} (4) | k=3 :{1,5,6,7} (4) | … | k=7 :{1,2,3,4,5} (5 ✓)

4STEP 4

The same holds for F, so {g(C), g(F)} = {1, 7}; by symmetry take g(C) = 1 and g(F) = 7.

{g(C), g(F)} = {1, 7}

5STEP 5

Grade 6 avoids F's neighbors so it lands on D; grade 2 avoids C's neighbors so it lands on G — g(D) = 6, g(G) = 2.

g(D) = 6, g(G) = 2

6STEP 6

E neighbors 1, 6, 7 force g(E) ∈ {3, 4}; g(E) = 3 leaves A and B as 4 and 5 (illegal on A-B), so g(E) = 4.

g(E) = 4

7STEP 7

Add the three requested grades.

g(C) + g(E) + g(F) = 1 + 4 + 7 = 12 → (A)

Answer

~12

A complete legal assignment g(C)=1, g(G)=2, g(B)=3, g(E)=4, g(A)=5, g(D)=6, g(F)=7 can be checked against all 12 walkways: every connected pair differs by at least 2 (the tightest being A-B with |5-3|=2 and C-D with |1-6|=5, all ≥ 2). The sum 1 + 4 + 7 = 12 matches choice (A). The opposite case g(C)=7, g(F)=1 gives the symmetric assignment with the same sum 7+4+1=12 — consistent and unique.

💡Key takeaway

This AMC 8 problem only needs Grade 6 absolute-value reasoning ("two grades must differ by at least 2") you already know!

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