2002 AMC 10A Problem 21

Attempt Problem 21 of the 2002 AMC 10A 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 10A solutions, or check the answer key.

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21.

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:meanmedian (data)modeextremal argument
Difficulty rating: 1660
Solution:

The sum is 88=64.8\cdot 8=64. 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.

If the largest were at least 16,16, the range condition would make the smallest at least 8.8. A mean of 88 would then force all eight integers to equal 8,8, contradicting the range.

If the largest were 15,15, the range 88 forces the smallest to be 7,7, so all eight integers are at least 7.7. The other seven then sum to 6415=49=77,64-15=49=7\cdot 7, forcing every one of them to equal 7.7. But then the median and mode would be 7,7, not 8,8, a contradiction.

Thus, the correct answer is D.

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