1997 AMC 8 第 14 题
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所有题目均经美国数学协会（MAA）官方合法授权使用。
14.
有一组五个正整数，它们的平均数是 ，中位数是 ，且唯一的众数是 。这组数中最大整数与最小整数的差是多少？
There is a set of five positive integers whose average (mean) is whose median is and whose only mode is What is the difference between the largest and smallest integers in the set?
小提示：
众数 必须恰好出现两次
The mode must appear exactly twice
大提示：
用平均数求出五个数的总和
Use the mean to get the total sum
解答：
五个数的和是 。
按从小到大排列，中间的数是 。因为 是唯一的众数，最后两个数都必须是 ；但不能有三个 ，否则中位数也会是 。
这三个已知数的和是 ，所以两个较小的数之和必须是 。它们必须是不同的正整数，否则另一个数也会与 并列为众数。因此它们是 和 。
因此所求的差为
所以正确答案是 D。
The sum of all the numbers in the list is
In increasing order, the middle number is . Because is the only mode, the last two numbers must both be ; there cannot be three s, since then the median would be .
These three known numbers sum to so the two smaller numbers must add to They must be distinct positive integers; otherwise a second value would tie as a mode. Thus they are and .
The desired difference is then
Thus, D is the correct answer.
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