2006 AIME II Problems/Problem 13
How many integers less than can be written as the sum of consecutive positive odd integers from exactly 5 values of ?
Let the first odd integer be , . Then the final odd integer is . The odd integers form an arithmetic sequence with sum . Thus, is a factor of .
Since , it follows that and .
Since there are exactly values of that satisfy the equation, there must be either or factors of . This means or . Unfortunately, we cannot simply observe prime factorizations of because the factor does not cover all integers for any given value of .
Instead we do some casework:
- If is odd, then must also be odd. For every odd value of , is also odd, making this case valid for all odd . Looking at the forms above and the bound of , must be
- Those give possibilities for odd .
- If is even, then must also be even. Substituting , we get
- Now we can just look at all the prime factorizations since cover the integers for any . Note that our upper bound is now :
- Those give possibilities for even .
The total number of integers is .
Let the largest odd number below the sequence be the th positive odd number, and the largest odd number in the sequence be the th positive odd number. Therefore, the sum is by sum of consecutive odd numbers. Note that and have the same parity, and can equal . We then perform casework based on the parity of .
If is odd, then must always be odd. Therefore, to have 5 pairs of odd factors, we must have either (in which case the number is a perfect square) or factors. Considering the upper bound, the only way this can happen is or . N must then be one of So, there are solutions when is odd.
If is even, then must have at least two factors of , so we can rewrite the expression as where . We can disregard the by dividing by and restricting our upper bound to . Since and don't have to be the same parity, we can include all cases less than 250 that have 9 or 10 factors. We then have as the possibilities.
Therefore, there are possibilities for ~sigma
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