2002 AMC 12A Problem 15

Below is the professionally curated solution for Problem 15 of the 2002 AMC 12A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2002 AMC 12A solutions, or check the answer key.

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Concepts:meanrangeextremal argument

Difficulty rating: 1660

15.

The mean, median, unique mode, and range of a collection of eight integers are all equal to 8.8. The largest integer that can be an element of this collection is

1111

1212

1313

1414

1515

Solution:

The collection 6,6,6,8,8,8,8,146, 6, 6, 8, 8, 8, 8, 14 has mean, median, unique mode, and range all equal to 8,8, so 1414 is attainable.

Suppose the largest were 15.15. The range 88 forces the smallest to be 7,7, and the median 88 fixes the two middle values as 8,8.8, 8. Then 7+8+8+15=38,7 + 8 + 8 + 15 = 38, so the remaining four values sum to 6438=26,64 - 38 = 26, averaging 6.5.6.5. At least one would be below 7,7, contradicting the minimum. So 1515 is impossible.

Thus, the correct answer is D.

Problem 15 in Other Years