AMC 8 · 2025 · #15

Grade 6 countingalgebra

Archetype Overlap / Complement Counting └ Sub-archetype Two-Attribute Table: Fill or Optimize a Cell

reflection-symmetrypaper-foldingsystems-of-equationscombinations-basic convert-to-algebraidentify-subproblems ↑ Prerequisites: linear-equations-two-varsystematic-enumeration

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A 6× 6 grid has 13 silver squares and 23 gold squares. Folding the grid vertically pairs column 1 with 6, 2 with 5, 3 with 4 — creating 18 overlapping pairs. Each pair is one of three types: silver-silver (SS), silver-gold (SG), or gold-gold (GG). Find the least value m and greatest value M of the number of GG pairs, then compute m+M.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

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

How to solve

Strategy Convert to Algebra

Counting silver tiles two ways links P_SS, P_SG, P_GG by two clean equations, and subtracting them gives the magic identity P_GG = P_SS + 5 (Tool #13). That single identity converts the original 'GG min/max' question into a 'SS min/max' question — Tool #16, count the complement — because gold-on-gold is hard to picture but silver-on-silver is a small, manageable count. Finally, Tool #7 splits the work into two parallel subproblems: find P_SS,min and P_SS,max independently, then convert each back to P_GG.

1STEP 1

Name the pair counts P_SS, P_SG, P_GG. Total pairs give one equation; counting silvers (2 per SS, 1 per SG) gives 2P_SS + P_SG = 13.

P_SS + P_SG + P_GG = 18 2P_SS + P_SG = 13

2STEP 2

Subtract the silver equation from the pair-total equation. P_SG cancels, leaving the tidy identity P_GG = P_SS + 5.

(P_SS + P_SG + P_GG) - (2P_SS + P_SG) = 18 - 13 → P_GG - P_SS = 5 → P_GG = P_SS + 5

3STEP 3

Since P_GG = P_SS + 5 always, optimizing P_GG means optimizing P_SS — and silvers are scarce, so SS pairs are far easier to count.

4STEP 4

Subproblem 1 (M): pack silvers into SS pairs. 13 silvers make at most ⌊ $\frac{13}{2}$ ⌋ = 6 SS pairs (1 silver left over), so M = 6 + 5 = 11.

P_SS,max = ⌊

\frac{13}{2}

⌋ = 6 → M = P_SS,max + 5 = 11

5STEP 5

Subproblem 2 (m): spread silvers so none overlap. With 18 slots and only 13 silvers, P_SS = 0 is easy, so m = 0 + 5 = 5.

P_SS,min = 0 → m = P_SS,min + 5 = 5

6STEP 6

Add the extremes: m + M = 5 + 11 = 16, which is choice (C).

m + M = 5 + 11 = 16 → (C)

Answer

Sanity-check the identity P_GG = P_SS + 5 at both extremes. When P_SS = 0: P_SG = 13, P_GG = 5, total 0 + 13 + 5 = 18. ✓ When P_SS = 6: P_SG = 13 - 12 = 1, P_GG = 11, total 6 + 1 + 11 = 18. ✓ All counts are non-negative integers ≤ 18, both colorings are realizable on the 6 × 6 grid, and m+M = 16 sits in the middle of the (A)-(E) range, which is plausible for an extremes-sum question.

💡Key takeaway

This AMC 8 problem only needs Grade 6 variable-and-equation skills — name the unknowns, subtract two count equations, and the answer pops out!

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