Difference between revisions of "2002 AMC 10A Problems/Problem 21"
(→Solution) |
|||
| Line 7: | Line 7: | ||
Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7. | Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7. | ||
| − | Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer <math>\boxed{\text{(D)}\ 14 }</math>. | + | Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer is <math>\boxed{\text{(D)}\ 14 }</math>. |
| + | |||
==See Also== | ==See Also== | ||
{{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}} | {{AMC10 box|year=2002|ab=A|num-b=20|num-a=22}} | ||
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
Revision as of 10:20, 27 December 2008
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
Since the mean is 8, the sum of the eight numbers must be 64. Now we use the fact that the range is 8 to experiment. Letting the smallest number be 7, we find the largest is 15. Since the unique mode is 8, there must be atleast two eights. The sum of those four numbers is 38 so the other four numbers sum to 26. That means the average of the other four numbers is 6.5, which is below 7. Hence, atleast one of the numbers is below 7, which contradicts our assumption that the smallest is 7.
Now let's try with the smallest number equal to 6. The largest number is then 14. If we take two sixes and four eights, we get a sequence that fits all the requirements. Hence, our answer is 
.
See Also
| 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 | ||
