Difference between revisions of "2012 AMC 8 Problems/Problem 22"
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Let the values of the missing integers be <math>x, y, z</math>. We will find the bound of the possible medians. | Let the values of the missing integers be <math>x, y, z</math>. We will find the bound of the possible medians. | ||
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The smallest possible median will happen when we order the set as <math>\{x, y, z, 2, 3, 4, 6, 9, 14\}</math>. The median is <math>3</math>. | The smallest possible median will happen when we order the set as <math>\{x, y, z, 2, 3, 4, 6, 9, 14\}</math>. The median is <math>3</math>. | ||
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The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math> | The largest possible median will happen when we order the set as <math>\{2, 3, 4, 6, 9, 14, x, y, z\}</math>. The median is <math>9</math> | ||
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Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, <math>\textbf{(D)}</math>. | Therefore, the median must be between <math>3</math> and <math>9</math> inclusive, yielding <math>7</math> possible medians, <math>\textbf{(D)}</math>. | ||
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~superagh | ~superagh |
Revision as of 18:37, 3 October 2020
Contents
Problem
Let be a set of nine distinct integers. Six of the elements are 2, 3, 4, 6, 9, and 14. What is the number of possible values of the median of ?
Solution 1
First, we find that the minimum value of the median of will be .
We then experiment with sequences of numbers to determine other possible medians.
Median:
Sequence:
Median:
Sequence:
Median:
Sequence:
Median:
Sequence:
Median:
Sequence:
Median:
Sequence:
Median:
Sequence:
Any number greater than also cannot be a median of set .
Therefore, the answer is
Solution 2
Let the values of the missing integers be . We will find the bound of the possible medians.
The smallest possible median will happen when we order the set as . The median is .
The largest possible median will happen when we order the set as . The median is
Therefore, the median must be between and inclusive, yielding possible medians, .
~superagh
See Also
2012 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 23 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.