AMC 10 · 2021 · #11

Grade 6 geometry-2d

Archetype Small-Domain Constrained Search

simons-favorite-factoring-trickfactorsarea-rectanglesoptimization-counting convert-to-algebrasystematic-enumeration ↑ Prerequisites: area-rectanglesfactors

📏 Medium solution 💡 3 insights

Problem

Grandma has just finished baking a large rectangular pan of brownies. She is planning to make rectangular pieces of equal size and shape, with straight cuts parallel to the sides of the pan. Each cut must be made entirely across the pan. Grandma wants to make the same number of interior pieces as pieces along the perimeter of the pan. What is the greatest possible number of brownies she can produce?

Pick an answer.

(A)

~24

(B)

~30

(C)

~48

(D)

~60

(E)

~64

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Toolkit + CCSS Solution

Understand

Restated: Grandma cuts a rectangular pan of brownies into equal rectangular pieces using cuts parallel to the sides. If $m$ pieces line one side and $n$ pieces line the other, the perimeter pieces are the ones on the outer border and the interior pieces are everything inside. She wants the count of perimeter pieces to equal the count of interior pieces. Find the greatest possible total number of pieces $mn$.

Givens: Cuts are parallel to the pan's sides, so the pieces form an $m \times n$ grid; Interior pieces form an $(m-2) \times (n-2)$ grid (requires $m \ge 3$ and $n \ge 3$); Perimeter pieces $=$ interior pieces; Answer choices: $24, 30, 48, 60, 64$

Unknowns: The greatest possible value of $mn$ (total pieces)

Understand

Plan

Primary tool: #1 Draw a Diagram

Secondary: #2 Make a Systematic List, #3 Eliminate Possibilities, #13 Convert to Algebra

Tool #1 (Diagram) makes the situation concrete: draw an $m \times n$ grid and shade the inside $(m-2) \times (n-2)$ block. From the picture the condition becomes $(m-2)(n-2) = 2(m+n) - 4$. Tool #13 (Algebra) rewrites this as $(m-4)(n-4) = 8$, a clean factor-pair equation. Tool #2 (Systematic List) enumerates the factor pairs of $8$ to get all $(m,n)$ candidates. Tool #3 (Eliminate) picks the candidate that maximizes $mn$ and confirms (D) against the answer choices.

Execute — Answer: D

#1 Draw a Diagram 3.MD.C.7 Step 1

Draw the pan as an $m \times n$ grid of unit pieces.
The pieces touching the outer edge form the perimeter band; the inner block of size $(m-2) \times (n-2)$ is the interior.
Count interior pieces $= (m-2)(n-2)$ and total pieces $= mn$, so perimeter pieces $= mn - (m-2)(n-2)$.

$$\text{interior} = (m-2)(n-2), \quad \text{perimeter} = mn - (m-2)(n-2)$$

💡 Grade 3 area-as-array: the inner rectangle's area counts the interior pieces.

#13 Convert to Algebra 6.EE.A.3 Step 2

Set perimeter pieces equal to interior pieces: $mn - (m-2)(n-2) = (m-2)(n-2)$.
So $mn = 2(m-2)(n-2)$.
Expand the right side: $2(m-2)(n-2) = 2mn - 4m - 4n + 8$.
Substitute back: $mn = 2mn - 4m - 4n + 8$.

$$mn = 2(m-2)(n-2) = 2mn - 4m - 4n + 8$$

💡 Grade 6 equivalent expressions: expanding makes the equation easier to rearrange.

#13 Convert to Algebra 6.EE.A.4 Step 3

Move everything to one side: $0 = mn - 4m - 4n + 8$, so $mn - 4m - 4n = -8$.
Add $16$ to both sides to complete the factoring: $mn - 4m - 4n + 16 = 8$, which factors as $(m-4)(n-4) = 8$.

$$(m-4)(n-4) = 8$$

💡 Grade 6 equivalent expressions: clever +16 turns a messy line into a product.

#2 Make a Systematic List 4.OA.B.4 Step 4

List positive integer factor pairs of $8$ (since $m, n \ge 3$, both factors $m-4$ and $n-4$ should be $\ge -1$; checking $0$ and $-1$ values shows they fail the divisibility setup, so only positive pairs give valid $m,n \ge 5$).
The positive pairs are $(1,8), (2,4), (4,2), (8,1)$.

$$(m-4, n-4) \in \{(1,8),(2,4),(4,2),(8,1)\}$$

💡 Grade 4 factor pairs: list all the ways $8$ splits into two whole-number factors.

#2 Make a Systematic List 3.OA.C.7 Step 5

Recover $(m, n)$ from each factor pair: $(5, 12), (6, 8), (8, 6), (12, 5)$.
Up to swapping, the two distinct pans are $5 \times 12$ and $6 \times 8$.
Their totals are $5 \cdot 12 = 60$ and $6 \cdot 8 = 48$.

$$(m, n) = (5,12) \Rightarrow mn = 60; \quad (m,n) = (6,8) \Rightarrow mn = 48$$

💡 Grade 3 multiplication facts: multiply each pair to get the total piece count.

#3 Eliminate Possibilities 4.OA.A.3 Step 6

Pick the larger total: $60 > 48$.
Verify the $5 \times 12$ pan: interior $= 3 \cdot 10 = 30$, perimeter $= 60 - 30 = 30$.
They match, so $5 \times 12$ is valid.
Match $60$ to the choices.

$$\max(mn) = 60 \Rightarrow \textbf{(D)}$$

💡 Grade 4 multi-step problem: compare options and match to the answer choices.

[1] #1 3.MD.C.7 Draw the pan as an $m \times n$ grid of unit pieces. The pieces touching the out

[2] #13 6.EE.A.3 Set perimeter pieces equal to interior pieces: $mn - (m-2)(n-2) = (m-2)(n-2)$. S

[3] #13 6.EE.A.4 Move everything to one side: $0 = mn - 4m - 4n + 8$, so $mn - 4m - 4n = -8$. Add

[4] #2 4.OA.B.4 List positive integer factor pairs of $8$ (since $m, n \ge 3$, both factors $m-4

[5] #2 3.OA.C.7 Recover $(m, n)$ from each factor pair: $(5, 12), (6, 8), (8, 6), (12, 5)$. Up t

[6] #3 4.OA.A.3 Pick the larger total: $60 > 48$. Verify the $5 \times 12$ pan: interior $= 3 \c

Review

Reasonableness: Double-check the $5 \times 12$ pan directly. Total pieces $= 60$. Border pieces: top row $12$ + bottom row $12$ + left column (without corners) $3$ + right column (without corners) $3$ = $30$. Interior pieces: $3 \cdot 10 = 30$. Border equals interior — exactly the condition. And $60$ is bigger than $6 \cdot 8 = 48$, so it's the maximum among valid pans. Magnitude is reasonable: a $5 \times 12$ pan of brownies is a real-world size, and $60$ pieces matches choice (D).

Alternative: Tool #3 (Eliminate Possibilities) plus Tool #6 (Guess and Check) directly on the answer choices. For each choice (total $T$), look for integer factorizations $T = mn$ with $(m-2)(n-2) = T/2$. For $T = 64$, you need $(m-2)(n-2) = 32$; the candidate $8 \times 8$ gives $36 \ne 32$, fail. For $T = 60$, the pan $5 \times 12$ gives $(3)(10) = 30 = 60/2$, valid. So (D) survives, others fall.

CCSS standards used (min grade 6)

3.MD.C.7 Relate area to multiplication and addition operations (Counting the interior pieces as the area $(m-2)(n-2)$ of the inner rectangle.)
3.OA.C.7 Fluently multiply and divide within 100 (Computing totals $5 \cdot 12 = 60$ and $6 \cdot 8 = 48$ for each pan.)
4.OA.A.3 Solve multi-step word problems using four operations with whole numbers (Comparing candidate totals and selecting the maximum against the answer choices.)
4.OA.B.4 Find all factor pairs and recognize multiples; determine prime or composite (Listing all positive-integer factor pairs of $8$ for $(m-4)(n-4)$.)
6.EE.A.3 Apply the properties of operations to generate equivalent expressions (Expanding $2(m-2)(n-2)$ and rearranging the equality.)
6.EE.A.4 Identify when two expressions are equivalent (Rewriting $mn - 4m - 4n + 16$ as $(m-4)(n-4)$.)

⭐ This AMC 10 problem only needs Grade 6 expression rewriting you already know! Set 'interior = perimeter' on an $m \times n$ pan: $(m-2)(n-2) = mn - (m-2)(n-2)$. A clean +16 trick turns it into $(m-4)(n-4) = 8$. The factor pairs of $8$ give pans $5 \times 12$ and $6 \times 8$, so the biggest total is $5 \cdot 12 = 60$, answer (D).

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: D

Review

CCSS standards used (min grade 6)

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