2009 AIME II Problems/Problem 7
Problem
Define to be for odd and for even. When is expressed as a fraction in lowest terms, its denominator is with odd. Find .
Solution
First, note that , and that .
We can now take the fraction and multiply both the numerator and the denominator by . We get that this fraction is equal to .
Now we can recognize that is simply , hence this fraction is , and our sum turns into .
Let . Obviously is an integer, and can be written as . Hence if is expressed as a fraction in lowest terms, its denominator will be of the form for some .
In other words, we just showed that . To determine , we need to determine the largest power of that divides .
Let be the largest such that that divides .
We can now return to the observation that . Together with the obvious fact that is odd, we get that .
It immediately follows that , and hence .
Obviously, for the function is is a strictly decreasing function. Therefore .
We can now compute . Hence .
And thus we have , and the answer is .
Additionally, once you count the number of factors of in the summation, one can consider the fact that, since must be odd, it has to take on a value of or (Because the number of s in the summation is clearly greater than , dividing by will yield a number greater than , and multiplying this number by any odd number greater than will yield an answer , which cannot happen on the AIME.) Once you calculate the value of , and divide by , must be equal to , as any other value of will result in an answer . This gives as the answer.
See Also
2009 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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