Difference between revisions of "2000 AMC 10 Problems/Problem 23"

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==Problem==
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#REDIRECT [[2000 AMC 12 Problems/Problem 14]]
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>?
 
 
 
<math>\mathrm{(A)}\ 3 \qquad\mathrm{(B)}\ 6 \qquad\mathrm{(C)}\ 9 \qquad\mathrm{(D)}\ 17 \qquad\mathrm{(E)}\ 20</math>
 
 
 
==Solution==
 
 
 
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>.
 
 
 
The sum of all numbers is <math>25+x</math>, therefore the mean is <math>\frac {25+x}7</math>.
 
 
 
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>.
 
 
 
We will now examine each of these three cases separately.
 
 
 
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.
 
 
 
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>.
 
 
 
Solving <math>2x-2 = \frac {25+x}7</math> we get the only solution for this case: <math>x=3</math>.
 
 
 
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>.
 
 
 
The sum of all solutions is therefore <math>3+17=\boxed{20}</math>.
 
 
 
==See Also==
 
 
 
{{AMC10 box|year=2000|num-b=22|num-a=24}}
 

Latest revision as of 00:06, 27 November 2011

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