2006 AIME II Problems/Problem 13
Problem
How many integers less than can be written as the sum of consecutive positive odd integers from exactly 5 values of ?
Solution
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 .
See also
2006 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
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