Difference between revisions of "2009 AIME II Problems/Problem 7"
(New page: == Problem == Define <math>n!!</math> to be <math>n(n-2)(n-4)\cdots 3\cdot 1</math> for <math>n</math> odd and <math>n(n-2)(n-4)\cdots 4\cdot 2</math> for <math>n</math> even. When <math>\...) |
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{{AIME box|year=2009|n=II|num-b=6|num-a=8}} | {{AIME box|year=2009|n=II|num-b=6|num-a=8}} | ||
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Revision as of 23:36, 4 July 2013
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 $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)}$ (Error compiling LaTeX. Unknown error_msg).
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.