2007 AIME I Problems/Problem 11
Problem
For each positive integer , let denote the unique positive integer such that . For example, and . If find the remainder when is divided by 1000.
Solution
and . Therefore if and only if is in this range, or . There are numbers in this range, so the sum of over this range is . , so all numbers to have their full range. Summing this up with the formula for the sum of the first squares (), we get . We need only consider the because we are working with modulo .
Now consider the range of numbers such that . These numbers are to . There are (1 to be inclusive) of them. , and , the answer.
See also
2007 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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