2000 AMC 10 Problem 23

Attempt Problem 23 of the 2000 AMC 10 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 10 solutions, or check the answer key.

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23.

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)arithmetic sequencecasework
Difficulty rating: 1950
Solution:

The mode is always 2,2, and the mean is μ=25+x7.\mu = \dfrac{25+x}{7}.

If x2,x \leq 2, the median is also 2.2. Two equal values cannot belong to a non-constant three-term arithmetic progression, so this case gives no solutions.

If 2<x<4,2 < x < 4, the values occur in the order 2<x<μ.2 < x < \mu. They form an arithmetic progression exactly when x2=μx.x-2 = \mu-x. Substituting for μ\mu gives x2=25+x7x, x-2 = \dfrac{25+x}{7}-x, whose solution is x=3.x=3.

If x4,x \geq 4, the median is 44 and μ>4.\mu>4. Thus the condition is 42=μ4, 4-2 = \mu-4, so μ=6\mu=6 and x=17.x=17.

The two possible values are therefore 33 and 17,17, whose sum is 20.20.

Thus, the correct answer is E.

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