2002 AMC 12A Problem 15

Attempt Problem 15 of the 2002 AMC 12A below, then check your answer against the professionally curated solution 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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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

Answer: D
Concepts:meanrangeextremal argument
Difficulty rating: 1660
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. Because 88 is the mode, it occurs in the sorted list; together with median 8,8, this forces the two middle values to be 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.

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