2007 iTest Problems/Problem 40
Problem
Let be the sum of all such that and . Compute .
Solution
Rewrite as . That results in Note that if is not an integer, than both sides can not equal each other. Thus, to find all solutions where , find the values where is an integer.
- The easier case is when is an integer. If is an integer, then , so both sides equal each other. Therefore, every integer from to is a solution.
- The harder case is when is not an integer. Let , where is an integer from to , and and are relatively prime integers, where . That means . Since and are relatively prime, must be a factor of . That means every mixed number in the form , where works.
With that taken in consideration, the sum equals The arithmetic series sum formula can be used to simplify things further. If the first term is , common difference is , and last term is , the sum of the terms in the series is . Now equals Thus, .
See Also
2007 iTest (Problems, Answer Key) | ||
Preceded by: Problem 39 |
Followed by: Problem 41 | |
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