2024 AMC 10B Problem 6

For a fixed area, a rectangle's perimeter is smallest when its two sides are as nearly equal as possible, so we want the factor pair of $2024$ closest to $\sqrt{2024}$.

Since $45^2 = 2025$, we have
$$2024 = 45^2 - 1 = (45 - 1)(45 + 1) = 44 \cdot 46.$$
No integer pair can be closer than two apart while multiplying to a non-square, so $44 \times 46$ is optimal. Its perimeter is
$$2(44 + 46) = 180.$$

The answer is $\boxed{\textbf{(B)}\ 180}$.

Among rectangles of fixed area, the square has the least perimeter (AM-GM: $l + w \ge 2\sqrt{lw}$), so integer problems reduce to finding the factor pair straddling the square root most tightly. Recognizing $2024$ as one less than a perfect square hands you that pair via the difference of squares. As a check, $2\sqrt{2024}$ is just under $90$, so the perimeter must be at least about $180$, which rules out (A).

• Picking a factor pair from the prime factorization 2^3 11 23 without checking that 44 and 46 are the closest pair, e.g. 23 * 88 giving 222.
• Reporting the semi-perimeter $44 + 46 = 90$ or the side sum instead of the full perimeter.