Difference between revisions of "2002 AMC 12A Problems/Problem 15"
(→Note) |
|||
| Line 32: | Line 32: | ||
The solution for <math>14</math> is, in fact, unique. As the median must be <math>8</math>, this means that both the <math>4^\text{th}</math> and the <math>5^\text{th}</math> number, when ordered by size, must be <math>8</math>s. This gives the partial solution <math>(6,a,b,8,8,c,d,14)</math>. For the mean to be <math>8</math> each missing variable must be replaced by the smallest allowed value. | The solution for <math>14</math> is, in fact, unique. As the median must be <math>8</math>, this means that both the <math>4^\text{th}</math> and the <math>5^\text{th}</math> number, when ordered by size, must be <math>8</math>s. This gives the partial solution <math>(6,a,b,8,8,c,d,14)</math>. For the mean to be <math>8</math> each missing variable must be replaced by the smallest allowed value. | ||
| + | One solution that works is <math>(6,6,6,8,8,8,8,14)</math> | ||
== See Also == | == See Also == | ||
Revision as of 19:36, 6 March 2019
- The following problem is from both the 2002 AMC 12A #15 and 2002 AMC 10A #21, so both problems redirect to this page.
Contents
Problem
The mean, median, unique mode, and range of a collection of eight integers are all equal to 8. The largest integer that can be an element of this collection is
Solution
As the unique mode is 
, there are at least two s.
As the range is and one of the numbers is , the largest one can be at most 
.
If the largest one is , then the smallest one is , and thus the mean is strictly larger than , which is a contradiction.
If the largest one is 
, then the smallest one is 
. This means that we already know four of the values: , , , . Since the mean of all the numbers is , their sum must be 
. Thus the sum of the missing four numbers is 
. But if is the smallest number, then the sum of the missing numbers must be at least 
, which is again a contradiction.
If the largest number is 
, we can easily find the solution 
. Hence, our answer is 
.
Note
The solution for is, in fact, unique. As the median must be , this means that both the 
and the 
number, when ordered by size, must be s. This gives the partial solution 
. For the mean to be each missing variable must be replaced by the smallest allowed value.
One solution that works is
See Also
| 2002 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 14 |
Followed by Problem 16 |
| 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 | |
| 2002 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 20 |
Followed by Problem 22 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
