AMC10 2002 A

AMC10 2002 A · Q21

AMC10 2002 A · Q21. It mainly tests Averages (mean), Arithmetic misc.

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is

一组八个整数的平均数、中位数、唯一众数和范围均为8。该集合中可能的最大整数是

(A) 11 11

(B) 12 12

(D) 14 14

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

(D) The values 6, 6, 6, 8, 8, 8, 8, 14 satisfy the requirements of the problem, so the answer is at least 14. If the largest number were 15, the collection would have the ordered form 7, __, __, 8, 8, __, __, 15. But $7+8+8+15=38$, and a mean of 8 implies that the sum of all values is 64. In this case, the four missing values would sum to $64-38=26$, and their average value would be 6.5. This implies that at least one would be less than 7, which is a contradiction. Therefore, the largest integer that can be in the set is 14.

(D) 数值 6、6、6、8、8、8、8、14 满足题目的要求，因此答案至少是 14。若最大数为 15，则该集合按从小到大的形式为 7，__，__，8，8，__，__，15。但 $7+8+8+15=38$，且平均数为 8 意味着所有数之和为 64。在这种情况下，四个缺失的数之和为 $64-38=26$，它们的平均值为 6.5。这意味着至少有一个数会小于 7，这与条件矛盾。因此，该集合中可能出现的最大整数是 14。

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