2002 AMC 12A Problems/Problem 15

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The following problem is from both the 2002 AMC 12A #15 and 2002 AMC 10A #21, so both problems redirect to this page.

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