2000 AMC 12 Problems/Problem 14

The following problem is from both the 2000 AMC 12 #14 and 2000 AMC 10 #23, so both problems redirect to this page.

Problem

When the mean, median, and mode of the list

\[10,2,5,2,4,2,x\]

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

$\text {(A)}\ 3 \qquad \text {(B)}\ 6 \qquad \text {(C)}\ 9 \qquad \text {(D)}\ 17 \qquad \text {(E)}\ 20$

Solution

  • The mean is $\frac{10+2+5+2+4+2+x}{7} = \frac{25+x}{7}$.
  • Arranged in increasing order, the list is $2,2,2,4,5,10$, so the median is either $2,4$ or $x$ depending upon the value of $x$.
  • The mode is $2$, since it appears three times.

We apply casework upon the median:

  • If the median is $2$ ($x \le 2$), then the arithmetic progression must be constant.
  • If the median is $4$ ($x \ge 4$), because the mode is $2$, the mean can either be $0,3,6$ to form an arithmetic progression. Solving for $x$ yields $-25,-4,17$ respectively, of which only $17$ works because it is larger than $4$.
  • If the median is $x$ ($2 \le x \le 4$), we must have the arithmetic progression $2, x, \frac{25+x}{7}$. Thus, we find that $2x=2+\frac{25+x}{7}$ so $x=3$.

The answer is $3 + 17 = 20\ \mathrm{(E)}$.

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
Preceded by
Problem 13
Followed by
Problem 15
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
2000 AMC 10 (ProblemsAnswer KeyResources)
Preceded by
Problem 22
Followed by
Problem 24
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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