Difference between revisions of "2009 AIME II Problems/Problem 7"
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It immediately follows that <math>p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)</math>, | It immediately follows that <math>p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)</math>, | ||
− | and hence <math>p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!) | + | and hence <math>p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)</math>. |
Obviously, for <math>i\in\{1,2,\dots,2009\}</math> the function <math>f(i)=2\cdot 2009 - i - p(i!)</math> is is a strictly decreasing function. | Obviously, for <math>i\in\{1,2,\dots,2009\}</math> the function <math>f(i)=2\cdot 2009 - i - p(i!)</math> is is a strictly decreasing function. |
Revision as of 22:47, 10 March 2015
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 denumerator 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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.