2004 AMC 12A 第 10 题

Question

49

个连续整数的和为

7^5

。它们的中位数是多少？

The sum of

49

consecutive integers is

7^5.

What is their median?

7

7^2

7^3

7^4

7^5

对连续整数来说，中位数等于平均数。

For consecutive integers the median equals the mean

中位数是总和除以项数，即

\dfrac{7^5}{49}

。

The median is the sum divided by the number of terms,

\dfrac{7^5}{49}

一组连续整数的和等于项数乘以平均数，而对连续整数来说，平均数等于中位数。

所以中位数为

\dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343\text{。}

所以正确答案是 C。

The sum of a set of consecutive integers equals the number of terms times their mean, and for consecutive integers the mean equals the median.

So the median is

\dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343.

Thus, the correct answer is C.

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