1989 AIME Problems/Problem 11
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
Let the mode be , which we let appear times. We let the arithmetic mean be , and the sum of the numbers be . Then As is essentially independent of , it follows that we wish to minimize or maximize (in other words, ). Indeed, is symmetric about ; consider replacing all of numbers in the sample with , and the value of remains the same. So, without loss of generality, let . Now, we would like to maximize the quantity
contains numbers that may appear at most times. Therefore, to maximize , we would have appear times, appear times, and so forth. We can thereby represent as the sum of arithmetic series of . We let , so
where denotes the sum of the remaining numbers, namely .
At this point, we introduce the crude estimate^{[1]} that , so and Expanding (ignoring the constants, as these do not affect which yields a maximum) and scaling, we wish to minimize the expression . By AM-GM, we have , with equality coming when , so . Substituting this result and some arithmetic gives an answer of .
In less formal language, it quickly becomes clear after some trial and error that in our sample, 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 .
Notes
- ^ In fact, when (which some simple testing shows that the maximum will occur around), it turns out that is an integer anyway, so indeed .
See also
1989 AIME (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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