Difference between revisions of "1989 AIME Problems/Problem 11"
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== Solution == | == Solution == | ||
It is obvious that there will be <math>n+1</math> values equal to one and <math>n</math> values each of <math>1000, 999, 998 \ldots</math>. It is fairly easy to find the [[maximum]]. Try <math>n=1</math>, which yields <math>924</math>, <math>n=2</math>, which yields <math>942</math>, <math>n=3</math>, which yields <math>947</math>, and <math>n=4</math>, which yields <math>944</math>. The maximum difference occurred at <math>n=3</math>, so the answer is <math>947</math>. | It is obvious that there will be <math>n+1</math> values equal to one and <math>n</math> values each of <math>1000, 999, 998 \ldots</math>. It is fairly easy to find the [[maximum]]. Try <math>n=1</math>, which yields <math>924</math>, <math>n=2</math>, which yields <math>942</math>, <math>n=3</math>, which yields <math>947</math>, and <math>n=4</math>, which yields <math>944</math>. The maximum difference occurred at <math>n=3</math>, so the answer is <math>947</math>. | ||
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== See also == | == See also == | ||
{{AIME box|year=1989|num-b=10|num-a=12}} | {{AIME box|year=1989|num-b=10|num-a=12}} | ||
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+ | [[Category:Intermediate Number Theory Problems]] |
Revision as of 19:45, 10 April 2008
Problem
A sample of 121 integers is given, each between 1 and 1000 inclusive, with repetitions allowed. The sample has a unique mode (most frequent value). Let be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of ? (For real , is the greatest integer less than or equal to .)
Solution
It is obvious that there will be values equal to one and values each of . It is fairly easy to find the maximum. Try , which yields , , which yields , , which yields , and , which yields . The maximum difference occurred at , so the answer is .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |