AMC 8 · 2025 · #15
Grade 6 countingalgebraProblem
Kei draws a 6-by-6 grid. He colors 13 of the unit squares silver and the remaining squares gold. Kei then folds the grid in half vertically, forming pairs of overlapping unit squares. Let m and M equal the least and greatest possible number of gold-on-gold pairs, respectively. What is the value of m+M?
Pick an answer.
AMC 8 2025 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: A $6\times 6$ 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 $m+M$.
Givens: Grid is $6 \times 6$, so $36$ unit squares total; Exactly $13$ squares are silver; the remaining $36 - 13 = 23$ are gold; Folding vertically creates $3 \times 6 = 18$ overlapping pairs; Each overlapping pair is SS, SG, or GG; Answer choices: (A) $12$, (B) $14$, (C) $16$, (D) $18$, (E) $20$
Unknowns: $m$ = least possible number of GG pairs; $M$ = greatest possible number of GG pairs; The value of $m + M$
Understand
Restated: A $6\times 6$ 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 $m+M$.
Givens: Grid is $6 \times 6$, so $36$ unit squares total; Exactly $13$ squares are silver; the remaining $36 - 13 = 23$ are gold; Folding vertically creates $3 \times 6 = 18$ overlapping pairs; Each overlapping pair is SS, SG, or GG; Answer choices: (A) $12$, (B) $14$, (C) $16$, (D) $18$, (E) $20$
Plan
Primary tool: #13 Convert to Algebra
Secondary: #7 Identify Subproblems, #16 Change Focus / Count the Complement
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 $P_{GG}$.
Execute — Answer: C
6.EE.B.6 Step 1 - Set up the bookkeeping.
- Let $P_{SS}$, $P_{SG}$, $P_{GG}$ be the number of pairs of each type.
- Total pairs gives one equation, and counting silver tiles (each SS pair contributes $2$ silvers, each SG pair contributes $1$) gives a second equation.
💡 Letting variables stand for unknown counts is exactly Grade 6 'use variables to write expressions for a problem'.
6.EE.A.3 Step 2 - Subtract the silver-tile equation from the pair-total equation.
- The $P_{SG}$ term cancels and we get a tidy identity linking the two same-color pair counts.
💡 Combining and simplifying two equivalent expressions to expose a hidden relationship is Grade 6 expression manipulation.
6.EE.A.4 Step 3 - Reframe the goal.
- Since $P_{GG} = P_{SS} + 5$ for every legal coloring, maximizing or minimizing $P_{GG}$ is the same as maximizing or minimizing $P_{SS}$.
- Counting silver-silver pairs is much easier than counting gold-gold pairs (silvers are scarce — only $13$ of them).
💡 Two expressions that always differ by $5$ are 'equivalent' for optimization purposes — minimize one, you minimize the other.
4.NBT.B.6 Step 4 - Subproblem 1 — find $M$ (maximum $P_{GG}$).
- Make as many SS pairs as possible.
- Each SS pair uses $2$ silver tiles, and we have $13$, so at most $\lfloor 13/2 \rfloor = 6$ SS pairs (leaving $1$ silver to form a single SG pair).
- Place all $6$ SS pairs by coloring both squares of each of $6$ chosen overlapping cells silver; the leftover silver goes in any other pair.
- This is achievable, so $P_{SS,\max} = 6$.
💡 Splitting $13$ into pairs with $1$ left over is exactly Grade 4 'quotient and remainder' thinking.
6.EE.B.5 Step 5 - Subproblem 2 — find $m$ (minimum $P_{GG}$).
- Make as few SS pairs as possible.
- We need $P_{SS} \ge 0$, and $P_{SS} = 0$ is achievable: scatter the $13$ silver tiles so that no two land in mirrored positions.
- With $18$ pair-slots and only $13$ silvers, this is easy (e.g.
- silvers in column $1$ only — its mirror column $6$ stays all gold).
💡 Checking that $P_{SS} = 0$ satisfies every constraint is Grade 6 'find values that make the equation true'.
4.OA.A.3 Step 6 - Add $m$ and $M$ to get the requested value.
- The result matches choice (C).
💡 The final assembly is a single Grade 4 multi-step word-problem sum.
6.EE.B.6 Set up the bookkeeping. Let $P_{SS}$, $P_{SG}$, $P_{GG}$ be the number of pairs 6.EE.A.3 Subtract the silver-tile equation from the pair-total equation. The $P_{SG}$ ter 6.EE.A.4 Reframe the goal. Since $P_{GG} = P_{SS} + 5$ for every legal coloring, maximizi 4.NBT.B.6 Subproblem 1 — find $M$ (maximum $P_{GG}$). Make as many SS pairs as possible. E 6.EE.B.5 Subproblem 2 — find $m$ (minimum $P_{GG}$). Make as few SS pairs as possible. We 4.OA.A.3 Add $m$ and $M$ to get the requested value. The result matches choice (C). Review
Reasonableness: 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 $\le 18$, both colorings are realizable on the $6 \times 6$ grid, and $m+M = 16$ sits in the middle of the (A)-(E) range, which is plausible for an extremes-sum question.
Alternative: Tool #16 (complement / focus shift) applied directly to GG pairs: of the $23$ gold tiles, count how many appear in SG pairs as $P_{SG}$, so $2P_{GG} + P_{SG} = 23$. Combined with $P_{SS} + P_{SG} + P_{GG} = 18$ this also gives $P_{GG} = P_{SS} + 5$. Or Tool #6 (Guess and Check): try $P_{SS}$ values $0, 1, 2, \ldots, 6$, read off $P_{GG} = P_{SS} + 5 \in \{5,\ldots,11\}$, sum the endpoints to get $5 + 11 = 16$.
CCSS standards used (min grade 6)
6.EE.B.6Use variables to represent numbers and write expressions to solve problems (Introducing $P_{SS}$, $P_{SG}$, $P_{GG}$ as the unknown pair counts and writing the two governing equations from the totals.)6.EE.A.3Apply the properties of operations to generate equivalent expressions (Subtracting one equation from the other to derive the identity $P_{GG} = P_{SS} + 5$.)6.EE.A.4Identify when two expressions are equivalent (Recognizing that minimizing/maximizing $P_{GG}$ is the same as minimizing/maximizing $P_{SS}$ because the two expressions always differ by $5$.)4.NBT.B.6Find whole-number quotients and remainders with up to four-digit dividends (Computing $\lfloor 13 / 2 \rfloor = 6$ with remainder $1$ to find the maximum number of silver-silver pairs.)6.EE.B.5Understand solving an equation or inequality as a process of finding values (Verifying that $P_{SS} = 0$ is an achievable value that satisfies all the pair-count constraints.)4.OA.A.3Solve multi-step word problems using four operations with whole numbers (Combining the two extreme values into the final sum $m + M = 5 + 11 = 16$.)
⭐ This AMC 8 problem only needs Grade 6 variable-and-equation skills — name the unknowns, subtract two count equations, and the answer pops out!
⭐ 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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