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 1
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.
Solution 2
Let be in the range of . Then, we need to find the point where the value of flips from to . This will happen when exceeds or . Thus, if then . For , then . There are terms in the first set of , and terms in the second set. Thus, the sum of from is or . For the time being, consider that . Then, the sum of the values of is . We can collapse this to . Now, we have to consider from . Considering from just , we see that all of these values have . Because there are values of in that range, the sum of in that range is . Adding this to we get or mod . Now, take the range . There are values of in this range, and each has . Thus, that contributes or to the sum. Finally, adding and we get .
~firebolt360
Solution 3
rearranging the inequalty, you get so the equation becomes from to testing the small values, values of result in , values of result in 2, ... so you can guess that values of result in the equation becomes
See also
2007 AIME I (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 |
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