2000 AMC 10 Problems/Problem 23

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 $x$ ?

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

Solution

As $2$ occurs three times and each of the three other values just once, regardless of what $x$ we choose the mode will always be $2$ .

The sum of all numbers is $25+x$ , therefore the mean is $\frac {25+x}7$ .

The six known values, in sorted order, are $2,2,2,4,5,10$ . From this sequence we conclude: If $x\leq 2$ , the median will be $2$ . If $2<x<4$ , the median will be $x$ . Finally, if $x\geq 4$ , the median will be $4$ .

We will now examine each of these three cases separately.

In the case $x\leq 2$ , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.

In the case $2<x<4$ we have $x < \frac {25+x}7$ , because $\frac {25+x}7 - x = \frac{25-6x}7 > \frac{25-6\cdot 4}7 > 0$ . Therefore our three values in order are $2,x,\frac {25+x}7$ . We want this to be an arithmetic progression. From the first two terms the difference must be $x-2$ . Therefore the third term must be $x+(x-2)=2x-2$ .

Solving $2x-2 = \frac {25+x}7$ we get the only solution for this case: $x=3$ .

The case $x\geq 4$ remains. Once again, we have $\frac {25+x}7 \geq \frac{25+4}7 > 4$ , therefore the order is $2,4,\frac {25+x}7$ . The only solution is when $6=\frac {25+x}7$ , i. e., $x=17$ .

The sum of all solutions is therefore $3+17=\boxed{20}$ .

2000 AMC 10 (Problems • Answer Key • Resources)
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
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2000 AMC 10 Problems/Problem 23

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