Difference between revisions of "2002 AMC 12A Problems/Problem 15"
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Since <math>f, g \ge 8</math>, <math>f+g \ge 16</math>. Plugging that in, we have <math>2a+b+c \le 24</math>. Note that we can't do the same for <math>b</math> and <math>c</math>, but we can do <math>b+c \ge 2a</math>, giving us <math>4a \le 24</math>, which means <math>a\le 6</math>. | Since <math>f, g \ge 8</math>, <math>f+g \ge 16</math>. Plugging that in, we have <math>2a+b+c \le 24</math>. Note that we can't do the same for <math>b</math> and <math>c</math>, but we can do <math>b+c \ge 2a</math>, giving us <math>4a \le 24</math>, which means <math>a\le 6</math>. | ||
We want to find <math>h=a+8\le 6+8=14</math> | We want to find <math>h=a+8\le 6+8=14</math> | ||
| − | This is our answer, so <math>\boxed{\text{(D)}14}</math> | + | This is our answer, so <math>\boxed{\text{(D) }14}</math> |
== See Also == | == See Also == | ||
Revision as of 21:24, 26 September 2021
- 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 1
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.
The solution that works is
Solution 2
Let the 8 numbers be 
, arranged in increasing order. Since the range of the eight numbers is 8, 
.
I claim that 
, 
must both be . Since the median is 8, the mean of and must be 8. Let's assume that and aren't . The mode of the collection is , and if and aren't, then must be between and (i.e. not in the collection). This is a contradiction, so and have to be 8.
Now, we have the eight numbers are 
.
Since the mean is , we have 
, giving us 
.
Since 
, 
. Plugging that in, we have 
. Note that we can't do the same for 
and 
, but we can do 
, giving us 
, which means 
.
We want to find 
This is our answer, so 
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. 
