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Difference between revisions of "1989 AIME Problems/Problem 11"
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== Solution ==
 
== Solution ==
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{{solution}}
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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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It is obvious that there will be n+1 values equal to one and n values each of 1000, 999, 998... .  It is fairly easy to find the maximum. Try n=1 (yields 924) n=2 (yields 942) n=3 (yields 947) and n=4 (yields 944). The maximum difference occered at n=3, so the answer is 947.
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== See also ==
 
== See also ==
 
* [[1989 AIME Problems/Problem 12|Next Problem]]
 
* [[1989 AIME Problems/Problem 12|Next Problem]]
 
* [[1989 AIME Problems/Problem 10|Previous Problem]]
 
* [[1989 AIME Problems/Problem 10|Previous Problem]]
 
* [[1989 AIME Problems]]
 
* [[1989 AIME Problems]]

Revision as of 17:47, 7 March 2007

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 $D^{}_{}$ be the difference between the mode and the arithmetic mean of the sample. What is the largest possible value of $\lfloor D^{}_{}\rfloor$? (For real $x^{}_{}$, $\lfloor x^{}_{}\rfloor$ is the greatest integer less than or equal to $x^{}_{}$.)

Solution

It is obvious that there will be $n+1$ values equal to one and $n$ values each of $1000, 999, 998 \ldots$. It is fairly easy to find the maximum. Try $n=1$, which yields $924$, $n=2$, which yields $942$, $n=3$, which yields $947$, and $n=4$, which yields $944$. The maximum difference occurred at $n=3$, so the answer is $947$.

See also

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