2024 AMC 10B Problem 6

1. Look at what makes a perimeter small. Try area $36$: a $1 \times 36$ rectangle has perimeter $74$, a $4 \times 9$ has $26$, and a $6 \times 6$ has $24$. The closer the two sides are, the smaller the perimeter.
2. So we want two whole numbers that multiply to $2024$ and are as close to each other as possible. The best would be two equal sides, each about the square root of $2024$.
3. Notice $45 \cdot 45 = 2025$, just one more than $2024$. Now move one step down and one step up from $45$:
$$44 \cdot 46 = 2024.$$
(Check: $44 \cdot 46 = 44 \cdot 45 + 44 = 1980 + 44 = 2024$.)
4. Could two sides be even closer? Closer would mean equal sides, and $2024$ is not a perfect square. So $44 \times 46$ is the best rectangle.
5. Its perimeter is
$$2 \cdot (44 + 46) = 2 \cdot 90 = 180.$$

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

For a fixed area, the perimeter is smallest when the rectangle is closest to a square. So the job is to find the factor pair nearest the square root. Here $2024$ is one less than $45^2$, and a number of the form $n^2 - 1$ always splits as $(n-1)(n+1)$.

Break $2024$ into primes: $2024 = 2 \cdot 2 \cdot 2 \cdot 11 \cdot 23$. To make two close factors, the smaller prime $11$ should get more of the $2$s: $11 \cdot 4 = 44$ and $23 \cdot 2 = 46$. The perimeter is $2(44 + 46) = 180$.

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 $\boxed{\textbf{(B)}\ 180}$.