Competition · AMC preparation · step 4 of 4
AMC 8 · 2025 · #21
Grade 6 logiccounting
Pick an answer.
AMC 8 2025 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
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.
List every walkway
Read the figure and list every connection — there are 12 walkways in all.
Reading a picture and sorting connections into a list is the same "classify and count" move kindergarteners do.
K.MD.B.3Draw A DiagramCount each pod's neighbors
Recount the walkways as a degree per pod; only C and F reach degree 5, so they are the most constrained.
Re-sorting the same information by "how many neighbors" lets us see who is most pinned down — still kindergarten counting.
K.MD.B.3Organize Information In More WaysRule out grades for degree 5
A degree-5 pod with grade k needs 5 usable grades for its neighbors; only k = 1 or k = 7 leaves that many.
Comparing |k - n| ≥ 2 across choices of k is exactly Grade 6 absolute-value reasoning.
A pod connected to five other pods can only be given the lowest grade or the highest grade.
▸ Why?
Once the pod takes a grade, its five neighbors are barred from that grade and from the grade just below and just above it; a middle grade blocks two neighbor grades and leaves only four free, but an end grade blocks just one and leaves five free.
▸ Why?
The seven grades split with no gaps or overlaps into the pod's own grade, the blocked neighbor grades, and the grades still free, so the free count is just seven with the blocked ones taken away.
▸ Why?
The five neighbors each need their own grade, so they are five distinct things dropped into the free grades acting as cells; a middle grade leaves only four cells, which forces two neighbors to land on the same grade — banned by the differ-by-two rule — while an end grade opens five cells so each neighbor gets its own.
Apply the same to pod F
The same holds for F, so {g(C), g(F)} = {1, 7}; by symmetry take g(C) = 1 and g(F) = 7.
After eliminating everything else, only the two extreme grades survive for both highly connected pods.
6.NS.C.7Eliminate PossibilitiesPlace grade 6
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.
Each extreme grade (2 next to 1, 6 next to 7) has exactly one legal home — the non-neighbor.
6.NS.C.7Eliminate PossibilitiesPin down pod E
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.
After eliminating each candidate using the A-B walkway as the tie-breaker, only one grade survives for E.
6.NS.C.7Eliminate PossibilitiesAdd the three grades
Add the three requested grades.
Adding three small whole numbers under 20 is Grade 1 arithmetic.
1.OA.A.2Eliminate PossibilitiesThis AMC 8 problem only needs Grade 6 absolute-value reasoning ("two grades must differ by at least 2") you already know!
- List every walkway
- Count each pod's neighbors
- Rule out grades for degree 5
- Apply the same to pod F
- Place grade 6
- Pin down pod E
- Add the three grades
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