AMC12 2020 A

AMC12 2020 A · Q8

AMC12 2020 A · Q8. It mainly tests Functions basics, Averages (mean).

What is the median of the following list of 4040 numbers? $1, 2, 3, \dots , 2020, 1^2, 2^2, 3^2, \dots , 2020^2$

以下 4040 个数的.median 是多少？ $1, 2, 3, \dots , 2020, 1^2, 2^2, 3^2, \dots , 2020^2$

(A) 1974.5 1974.5

(B) 1975.5 1975.5

(D) 1977.5 1977.5

(E) 1978.5 1978.5

Answer

Correct choice: (C)

正确答案：（C）

Solution

Because there are 4040 numbers in the list, the median is the mean of the numbers in positions 2020 and 2021 when the list is put into nondecreasing order. All of the squared entries are greater than all of the non-squared entries except for $1^2, 2^2, \dots, 44^2 = 1936$. When the numbers are put into nondecreasing order, those 44 entries will be in the first half, and the greatest 44 non-squared entries, namely 1977, 1978, $\dots$, 2020, will be in the second half. Therefore the entry in position 2020 will be 1976, the entry in position 2021 will be 1977, and the median will be 1976.5.

因为列表中有 4040 个数，中位数是列表按非递减顺序排列时位置 2020 和 2021 的数的平均值。除了 $1^2, 2^2, \dots, 44^2 = 1936$ 外，所有平方项都大于所有非平方项。将数按非递减顺序排列时，这 44 个项将在前半部分，最大的 44 个非平方项，即 1977、1978、$\dots$、2020，将在后半部分。因此位置 2020 的数是 1976，位置 2021 的数是 1977，中位数是 1976.5。

Topics