AMC12 2002 A

AMC12 2002 A · Q15

AMC12 2002 A · Q15. It mainly tests Averages (mean), Casework.

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

由八个整数构成的一组数，其平均数、中位数、唯一众数和极差都等于 8。该组数中可能出现的最大整数是

(A) 11 11

(B) 12 12

(D) 14 14

(E) 15 15

Answer

Correct choice: (D)

正确答案：（D）

Solution

As the unique mode is $8$, there are at least two $8$s. As the range is $8$ and one of the numbers is $8$, the largest one can be at most $16$. If the largest one is $16$, then the smallest one is $8$, and thus the mean is strictly larger than $8$, which is a contradiction. If we have 2 8's we can find the numbers 4, 6, 7, 8, 8, 9, 10, 12. This is a possible solution but has not reached the maximum. If we have 4 8's we can find the numbers 6, 6, 6, 8, 8, 8, 8, 14. We can also see that they satisfy the need for the mode, median, and range to be 8. This means that the answer will be $\boxed{\text{(D)}\ 14 }$. ~By QWERTYUIOPASDFGHJKLZXCVBNM

由于唯一众数是 $8$，所以至少有两个 $8$。由于极差为 $8$ 且其中一个数是 $8$，最大值至多为 $16$。若最大值为 $16$，则最小值为 $8$，从而平均数严格大于 $8$，矛盾。若有两个 $8$，可以取 4, 6, 7, 8, 8, 9, 10, 12。这是一个可行解，但还未达到最大值。若有四个 $8$，可以取 6, 6, 6, 8, 8, 8, 8, 14。也可看出它们满足众数、中位数与极差都为 8 的要求。因此答案为 $\boxed{\text{(D)}\ 14 }$. ~By QWERTYUIOPASDFGHJKLZXCVBNM

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