AMC 10 · 2021 · #11

Grade 6 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

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

📏 Medium solution 💡 3 insights

Problem

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.

Pick an answer.

(A)

~24

(B)

~30

(C)

~48

(D)

~60

(E)

~64

AMC 10 2021 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Tool #1 (Diagram) makes the situation concrete: draw an m × n grid and shade the inside (m-2) × (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.

1STEP 1

Draw the m × n grid: the inner (m-2) × (n-2) block is the interior, so interior = (m-2)(n-2) and perimeter = mn - (m-2)(n-2).

interior = (m-2)(n-2), perimeter = mn - (m-2)(n-2)

2STEP 2

Set perimeter = interior: mn - (m-2)(n-2) = (m-2)(n-2), so mn = 2(m-2)(n-2) = 2mn - 4m - 4n + 8.

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

3STEP 3

Bring all to one side, then add 16 (Simon's trick) to factor: mn - 4m - 4n + 16 = 8, i.e. (m-4)(n-4) = 8.

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

4STEP 4

Since m, n ≥ 3, both factors are positive; list the factor pairs of 8: (1,8), (2,4), (4,2), (8,1).

(m-4, n-4) ∈ {(1,8),(2,4),(4,2),(8,1)}

5STEP 5

Recover (m, n): the two distinct pans are 5 × 12 and 6 × 8, with totals 5 · 12 = 60 and 6 · 8 = 48.

(m, n) = (5,12) → mn = 60; (m,n) = (6,8) → mn = 48

6STEP 6

Larger total 60 ＞ 48 wins; check 5 × 12: interior = 3 · 10 = 30 = perimeter 60 - 30, so the max is (D).

max(mn) = 60 → (D)

Answer

~60

Double-check the 5 × 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 · 10 = 30. Border equals interior — exactly the condition. And 60 is bigger than 6 · 8 = 48, so it's the maximum among valid pans. Magnitude is reasonable: a 5 × 12 pan of brownies is a real-world size, and 60 pieces matches choice (D).

💡Key takeaway

This AMC 10 problem only needs Grade 6 expression rewriting you already know! Set 'interior = perimeter' on an m × 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 × 12 and 6 × 8, so the biggest total is 5 · 12 = 60, answer (D).

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