AMC10 2024 B
AMC10 2024 B · Q6
AMC10 2024 B · Q6. It mainly tests Primes & prime factorization, Counting divisors.
A rectangle has integer length sides and an area of 2024. What is the least possible perimeter of the rectangle?
一个长宽为整数的矩形，面积为2024。求该矩形的最小可能周长？
(A)
160
160
(B)
180
180
(C)
222
222
(D)
228
228
(E)
390
390
Answer
Correct choice: (B)
正确答案：（B）
Solution
We can start by assigning the values x and y for both sides. Here is the equation representing the area:
$x \cdot y = 2024$
Let's write out 2024 fully factorized.
$2^3 \cdot 11 \cdot 23$
Since we know that $x^2 > (x+1)(x-1)$, we want the two closest numbers possible. After some quick analysis, those two numbers are $44$ and $46$. $\\44+46=90$
Now we multiply by $2$ and get $\boxed{\textbf{(B) }180}.$
Solution by IshikaSaini.
我们用 $x$ 和 $y$ 表示两边长度。面积方程为：
$x \cdot y = 2024$
将2024完全因式分解：
$2^3 \cdot 11 \cdot 23$
由于 $x^2 > (x+1)(x-1)$，我们希望两边尽可能接近。经过快速分析，这两个数是 $44$ 和 $46$。$44+46=90$
周长乘以 $2$ 得到 $\boxed{\textbf{(B) }180}$。
Solution by IshikaSaini.
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