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

(Solution)
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We apply casework upon the median:
 
We apply casework upon the median:
 
*If the median is <math>2</math> (<math>x \le 2</math>), then the arithmetic progression must be constant.
 
*If the median is <math>2</math> (<math>x \le 2</math>), then the arithmetic progression must be constant.
*If the median is <math>4</math> (<math>x \ge 4</math>), because the mode is <math>2</math>, the mean can either be <math>0,3,6</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-25,-4,17</math> respectively, of which only <math>17</math> works because it is larger than <math>14</math>.
+
*If the median is <math>4</math> (<math>x \ge 4</math>), because the mode is <math>2</math>, the mean can either be <math>0,3,6</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-25,-4,17</math> respectively, of which only <math>17</math> works because it is larger than <math>4</math>.
 
*If the median is <math>x</math> (<math>2 \le x \le 4</math>), again, because the mode is <math>2</math>, the mean can either be <math>1, 5/2, 4</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-18,-7.5, 3</math> respectively, of which only <math>3</math> works because it is greater than <math>2</math>, and less than <math>4</math>.
 
*If the median is <math>x</math> (<math>2 \le x \le 4</math>), again, because the mode is <math>2</math>, the mean can either be <math>1, 5/2, 4</math> to form an arithmetic progression. Solving for <math>x</math> yields <math>-18,-7.5, 3</math> respectively, of which only <math>3</math> works because it is greater than <math>2</math>, and less than <math>4</math>.
  

Revision as of 11:20, 24 May 2018

The following problem is from both the 2000 AMC 12 #14 and 2000 AMC 10 #23, so both problems redirect to this page.

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 $x$?

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

Solution

  • 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.
  • If the median is $4$ ($x \ge 4$), because the mode is $2$, 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 because it is larger than $4$.
  • If the median is $x$ ($2 \le x \le 4$), again, because the mode is $2$, the mean can either be $1, 5/2, 4$ to form an arithmetic progression. Solving for $x$ yields $-18,-7.5, 3$ respectively, of which only $3$ works because it is greater than $2$, and less than $4$.

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

See also

2000 AMC 12 (ProblemsAnswer KeyResources)
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
2000 AMC 10 (ProblemsAnswer KeyResources)
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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