Difference between revisions of "2000 AMC 10 Problems/Problem 23"
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==Problem== | ==Problem== | ||
+ | When the mean, median, and mode of the list <cmath>10,2,5,2,4,2,x</cmath> are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of <math>x</math>? | ||
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+ | <math>\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 17 \qquad\mathrm{(E)}\ 20</math> | ||
==Solution== | ==Solution== | ||
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+ | As <math>2</math> occurs three times and each of the three other values just once, regardless of what <math>x</math> we choose the mode will always be <math>2</math>. | ||
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+ | The sum of all numbers is <math>25+x</math>, therefore the mean is <math>\frac {25+x}7</math>. | ||
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+ | The six known values, in sorted order, are <math>2,2,2,4,5,10</math>. From this sequence we conclude: If <math>x\leq 2</math>, the median will be <math>2</math>. If <math>2<x<4</math>, the median will be <math>x</math>. Finally, if <math>x\geq 4</math>, the median will be <math>4</math>. | ||
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+ | We will now examine each of these three cases separately. | ||
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+ | In the case <math>x\leq 2</math>, both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression. | ||
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+ | In the case <math>2<x<4</math> we have <math>x < \frac {25+x}7</math>, because <math>\frac {25+x}7 - x = \frac{25-6x}7 > \frac{25-6\cdot 4}7 > 0</math>. | ||
+ | Therefore our three values in order are <math>2,x,\frac {25+x}7</math>. We want this to be an arithmetic progression. From the first two terms the difference must be <math>x-2</math>. Therefore the third term must be <math>x+(x-2)=2x-2</math>. | ||
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+ | Solving <math>2x-2 = \frac {25+x}7</math> we get the only solution for this case: <math>x=3</math>. | ||
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+ | The case <math>x\geq 4</math> remains. Once again, we have <math>\frac {25+x}7 \geq \frac{25+4}7 > 4</math>, therefore the order is <math>2,4,\frac {25+x}7</math>. The only solution is when <math>6=\frac {25+x}7</math>, i. e., <math>x=17</math>. | ||
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+ | The sum of all solutions is therefore <math>3+17=\boxed{20}</math>. | ||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2000|num-b=22|num-a=24}} | {{AMC10 box|year=2000|num-b=22|num-a=24}} |
Revision as of 10:28, 11 January 2009
Problem
When the mean, median, and mode of the list are arranged in increasing order, they form a non-constant arithmetic progression. What is the sum of all possible real values of ?
Solution
As occurs three times and each of the three other values just once, regardless of what we choose the mode will always be .
The sum of all numbers is , therefore the mean is .
The six known values, in sorted order, are . From this sequence we conclude: If , the median will be . If , the median will be . Finally, if , the median will be .
We will now examine each of these three cases separately.
In the case , both the median and the mode are 2, therefore we can not get any non-constant arithmetic progression.
In the case we have , because . Therefore our three values in order are . We want this to be an arithmetic progression. From the first two terms the difference must be . Therefore the third term must be .
Solving we get the only solution for this case: .
The case remains. Once again, we have , therefore the order is . The only solution is when , i. e., .
The sum of all solutions is therefore .
See Also
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 | ||
All AMC 10 Problems and Solutions |