2000 AMC 12 Problem 14

Attempt Problem 14 of the 2000 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AMC 12 solutions, or check the answer key.

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14.

When the mean, median, and mode of the list 10,2,5,2,4,2,x10, 2, 5, 2, 4, 2, x

are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of x?x?

33

66

99

1717

2020

Answer: E
Concepts:meanmedian (data)modecasework
Difficulty rating: 1840
Solution:

The six fixed numbers sum to 25,25, so the mean is 25+x7,\dfrac{25 + x}{7}, and the mode is 2.2. If x2,x \le 2, then 22 is both median and mode, forcing a constant progression, so x>2.x \gt 2.

Case 2<x<42 \lt x \lt 4: the median is x.x. Requiring 2,x,25+x72, x, \dfrac{25 + x}{7} to form an arithmetic progression yields x=3x = 3 as the only value in this range.

Case x4x \ge 4: the median is 4,4, and the progression 2,4,62, 4, 6 forces the mean to be 6,6, so 25+x7=6,\dfrac{25 + x}{7} = 6, giving x=17.x = 17.

The sum of all possible values is 3+17=20.3 + 17 = 20.

Thus, the correct answer is E.

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