Difference between revisions of "2000 AMC 12 Problems/Problem 14"

Revision as of 19:03, 4 January 2008

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 $x$ ?

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

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, which results in a contradiction.
If the median is $4$ ( $x \ge 4$ ), then 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.
If the median is $x$ ( $2 \le x \le 4$ ), then the mean can either be $2-(x-2), \frac{x+2}{2}, 2+(x-2)$ . Solving for $x$ yields $\frac{3}{8}, \frac{36}{5}, 3$ respectively, of which only $3$ works.

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

2000 AMC 12 (Problems • Answer Key • Resources)
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

@@ Line 21: / Line 21: @@
 == See also ==
-{{AMC12 box|year=2000|num-b=8|num-a=10}}
+{{AMC12 box|year=2000|num-b=13|num-a=15}}
 [[Category:Introductory Algebra Problems]]