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

(New page: ==Problem== ==Solution== ==See Also== {{AMC10 box|year=2000|num-b=22|num-a=24}})
 
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==Problem==
 
==Problem==
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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>.
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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 \[10,2,5,2,4,2,x\] 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 $3+17=\boxed{20}$.

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
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