AMC 10 · 2024 · #6

Grade 8 geometry-2d
factorsprime-factorizationperimeterarea-rectangles systematic-enumerationidentify-subproblems ↑ Prerequisites: prime-factorizationfactorsperimeterarea-rectangles
📏 Medium solution 💡 2 insights
Problem
A rectangle has whole-number side lengths and area 2024. Among all such rectangles, find the smallest possible perimeter.

Pick an answer.

(A)
160
(B)
180
(C)
222
(D)
228
(E)
390

AMC 10 2024 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Guess and Check

For a fixed product, two positive numbers whose sum is smallest are the pair closest to the square root. Since √(2024) ≈ 45, we only need to test integers near 45 — a perfect setup for Tool #6 (Guess and Check) starting from the square-root estimate and adjusting up or down. Tool #7 (Identify Subproblems) splits the work into two pieces — first find the prime factorization of 2024 so we know which integers near 45 are actually divisors, then turn the winning (ℓ, w) pair into a perimeter.

1STEP 1

Subproblem A: prime-factor 2024 = 2³ × 11 × 23, so every side length is built from these primes.

2024 = 2 × 1012 = 2 × 2 × 506 = 2³ × 253 = 2³ × 11 × 23
2STEP 2

Estimate √2024 ≈ 45 (since 45² = 2025) to know where the balanced factor pair should sit.

44² = 1936, 45² = 2025 → √(2024) ≈ 45
3STEP 3

Guess near 45: split the primes into 46 × 44 = 2024, with both sides just 1 away from 45.

46 × 44 = 2024 ✓
4STEP 4

Check the neighbor: the factor just below 44 is 23, pairing with 88 for sum 111 — above 90, so (44, 46) wins.

next pair: 23 × 88 → 23 + 88 = 111 > 90
5STEP 5

Subproblem B: plug (46, 44) into P = 2(ℓ + w) = 2 × 90 = 180 → choice (B).

P = 2(46 + 44) = 2 × 90 = 180 → (B)
Answer
180
Eyeball the answer choices. Choice (A) 160 would need ℓ + w = 80, meaning a pair averaging 40, but 40 × 40 = 1600 < 2024, so the product can't reach 2024 — (A) is impossible. Our answer 180 corresponds to ℓ + w = 90, average 45, matching √(2024) ≈ 45 exactly. (C), (D), (E) correspond to lopsided pairs like 23 × 88 or 11 × 184, all larger.
💡Key takeaway

This AMC 10 problem only needs Grade 8 square-root estimation you already know — find the pair of factors closest to √(area) and the perimeter is forced!