2002 AMC 10A Problems/Problem 21
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.
See Also
2002 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 20 |
Followed by Problem 22 | |
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All AMC 10 Problems and Solutions |