AMC 10 · 2024 · #6
Grade 8 geometry-2dProblem
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
Pick an answer.
AMC 10 2024 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 rectangle has whole-number side lengths and area $2024$. Among all such rectangles, find the smallest possible perimeter.
Givens: Both side lengths are positive integers; Area $= \text{length} \times \text{width} = 2024$; Perimeter $= 2(\text{length} + \text{width})$; Answer choices: (A) $160$, (B) $180$, (C) $222$, (D) $228$, (E) $390$
Unknowns: The least possible perimeter, in the same length units as the sides
Understand
Restated: A rectangle has whole-number side lengths and area $2024$. Among all such rectangles, find the smallest possible perimeter.
Givens: Both side lengths are positive integers; Area $= \text{length} \times \text{width} = 2024$; Perimeter $= 2(\text{length} + \text{width})$; Answer choices: (A) $160$, (B) $180$, (C) $222$, (D) $228$, (E) $390$
Plan
Primary tool: #6 Guess and Check
Secondary: #7 Identify Subproblems
For a fixed product, two positive numbers whose sum is smallest are the pair closest to the square root. Since $\sqrt{2024} \approx 45$, 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 $(\ell, w)$ pair into a perimeter.
Execute — Answer: B
6.NS.B.4 Step 1 - Subproblem A: factor $2024$ so we know which integer side lengths are even allowed.
- Divide out the factor of $2$ as many times as possible, then test small primes on what remains.
💡 Prime factorization lists every "building block" of $2024$, so every divisor is built by choosing some of these primes — Grade 6 "factors and multiples".
8.EE.A.2 Step 2 - Estimate the square root of $2024$ to know where to start guessing.
- The closest perfect square is $45^2 = 2025$, so $\sqrt{2024}$ is just under $45$.
💡 The closer the two sides are, the smaller $\ell + w$. So the best pair will sit on either side of $45$ — Grade 8 square-root estimation.
4.OA.B.4 Step 3 - Guess and check near $45$.
- Combine the prime factors $\{2,2,2,11,23\}$ into two products as balanced as possible.
- Try $\ell = 2 \times 23 = 46$ and $w = 2 \times 2 \times 11 = 44$ — both within $1$ of $45$.
💡 Trying numbers right next to $\sqrt{2024}$ first is the Tool #6 "educated guess" move — Grade 4 factor-pair thinking.
4.OA.B.4 Step 4 - Confirm no closer pair exists.
- The next factor below $44$ is $23$, paired with $88$ — a sum of $111$, much bigger than $44 + 46 = 90$.
- So $(44, 46)$ wins.
💡 Once we step away from $\sqrt{2024}$, the pair spreads apart fast — a quick "check the neighbor" rules it out.
4.MD.A.3 Step 5 - Subproblem B: turn the minimizing pair into a perimeter.
- Apply $P = 2(\ell + w)$ with $(\ell, w) = (46, 44)$.
💡 The rectangle perimeter formula is Grade 4 measurement — just plug the winning sides in.
6.NS.B.4 Subproblem A: factor $2024$ so we know which integer side lengths are even allow 8.EE.A.2 Estimate the square root of $2024$ to know where to start guessing. The closest 4.OA.B.4 Guess and check near $45$. Combine the prime factors $\{2,2,2,11,23\}$ into two 4.OA.B.4 Confirm no closer pair exists. The next factor below $44$ is $23$, paired with $ 4.MD.A.3 Subproblem B: turn the minimizing pair into a perimeter. Apply $P = 2(\ell + w)$ Review
Reasonableness: Eyeball the answer choices. Choice (A) $160$ would need $\ell + w = 80$, meaning a pair averaging $40$, but $40 \times 40 = 1600 < 2024$, so the product can't reach $2024$ — (A) is impossible. Our answer $180$ corresponds to $\ell + w = 90$, average $45$, matching $\sqrt{2024} \approx 45$ exactly. (C), (D), (E) correspond to lopsided pairs like $23 \times 88$ or $11 \times 184$, all larger.
Alternative: Tool #3 (Eliminate Possibilities) on the answers: each choice $P$ forces $\ell + w = P/2$. Combined with $\ell w = 2024$, the pair $(\ell, w)$ satisfies $x^2 - (P/2)x + 2024 = 0$. The discriminant $(P/2)^2 - 4 \times 2024$ must be a nonnegative perfect square for integer sides. (A) gives $80^2 - 8096 = -1696 < 0$ — impossible. (B) gives $90^2 - 8096 = 4 = 2^2$ — works, with sides $44, 46$. So (B) is the smallest valid perimeter.
CCSS standards used (min grade 8)
6.NS.B.4Find greatest common factor and least common multiple of two numbers (Building the prime factorization $2024 = 2^3 \times 11 \times 23$ so that every possible integer side length can be read off as a product of some subset of these primes.)8.EE.A.2Use square root and cube root symbols to represent solutions (Estimating $\sqrt{2024} \approx 45$ from $44^2 = 1936$ and $45^2 = 2025$ to locate where the best factor pair must live.)4.OA.B.4Find all factor pairs and recognize multiples; determine prime or composite (Choosing the factor pair $(44, 46)$ closest to $\sqrt{2024}$ as the candidate that minimizes $\ell + w$, and ruling out the neighboring pair $(23, 88)$.)4.MD.A.3Apply area and perimeter formulas for rectangles in real-world problems (Computing the rectangle's perimeter $P = 2(\ell + w) = 2(44 + 46) = 180$ from the winning side lengths.)
⭐ This AMC 10 problem only needs Grade 8 square-root estimation you already know — find the pair of factors closest to $\sqrt{\text{area}}$ and the perimeter is forced!
⭐ This AMC 10 problem only needs Grade 8 square-root estimation you already know — find the pair of factors closest to $\sqrt{\text{area}}$ and the perimeter is forced!
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