AMC12 2004 A
AMC12 2004 A · Q10
AMC12 2004 A · Q10. It mainly tests Functions basics, Averages (mean).
The sum of $49$ consecutive integers is $7^5$. What is their median?
$49$ 个连续整数的和为 $7^5$。它们的中位数是多少？
(A)
7
7
(B)
72
72
(C)
73
73
(D)
74
74
(E)
75
75
Answer
Correct choice: (C)
正确答案：（C）
Solution
The median of a sequence is the middle number of the sequence when the sequence is arranged in order. Since the integers are consecutive, the median is also the mean, so the median is $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$.
一个序列的中位数是将序列按顺序排列后位于中间的数。由于这些整数是连续的，中位数也等于平均数，因此中位数为 $\frac{7^5}{49} = 7^3\ \mathrm{(C)}$。
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