2009 AIME II Problems/Problem 7
Contents
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 1
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.
Just a small note. It's important to note the properties of the function, which is what Solution 1 is using but denoting it as
.
We want to calculate
as the final step. We know that one property of
is that
.
Therefore, we have that
. Thus, we see by similar calculations as in Solution 1, that
. From which the conclusion follows.
- (OmicronGamma)
Solution 2
Using the steps of the previous solution we get and if you do the small cases(like
) you realize that youu can "thin-slice" the problem and simply look at the cases where
(they're nearly identical in nature but one has
with it) since
hardly contains any powers of
or in other words it's very inefficient and the inefficient cases hold all the power so you can just look at the highest powers of
in
and
and you get the minimum power of
in either expression is
so the answer is
since it would violate the rules of the AIME and the small cases if
.
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.