2020 AIME II Problems/Problem 8
Define a sequence recursively by and for integers . Find the least value of such that the sum of the zeros of exceeds .
Solution 1 (Official MAA)
First it will be shown by induction that the zeros of are the integers , where
This is certainly true for . Suppose that it is true for , and note that the zeros of are the solutions of , where is a nonnegative zero of . Because the zeros of form an arithmetic sequence with common difference so do the zeros of . The greatest zero of isso the greatest zero of is and the least is .
It follows that the number of zeros of is , and their average value is . The sum of the zeros of isLet , so the sum of the zeros exceeds if and only if Because is increasing for , the values and show that the requested value of is .
Solution 2 (Same idea, easier to see)
Starting from , we can track the solutions, the number of solutions, and their sum.
It is clear that the solutions form arithmetic sequences with a difference of 2, and the negative solutions cancel out all but of the solutions. Thus, the sum of the solutions is , which is a cubic function.
Multiplying both sides by ,
1 million is , so the solution should be close to .
100 is slightly too small, so works.
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