Difference between revisions of "2009 AIME II Problems/Problem 7"
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== Solution 3 (Divisibility) == | == Solution 3 (Divisibility) == | ||
We can logically deduce that the <math>b</math> value will be 1. Listing out the first few values of odd and even integers, we have: | We can logically deduce that the <math>b</math> value will be 1. Listing out the first few values of odd and even integers, we have: | ||
− | <math>1, 3, 5, 7, 9, 11, 13</math> and <math>2, 4, 6, 8, 10, 12, 14, 16</math>. Obviously, none of the factors of <math>2</math> in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of <math>3</math> occurs every <math>3</math> terms, and starting out on the third term of the denominator, a factor of <math>3</math> appears also every <math>3</math> terms. Thus, the factors of <math>3</math> on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form <math>1 \cdot 2^a</math>. Since there will be no odd factors in the denominators, all the denominators will be in the form <math>2^a</math> where <math>a</math> is the number of factors of <math>2</math> in <math>(2009 \cdot 2)! = 4018!</math>. This is simply <math>\sum_{n=1}^{11} \left \lfloor{\frac{4018}{2^n}}\right \rfloor = 4010</math>. Therefore, our answer is <math>\boxed{ | + | <math>1, 3, 5, 7, 9, 11, 13...</math> and <math>2, 4, 6, 8, 10, 12, 14, 16...</math>. Obviously, none of the factors of <math>2</math> in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of <math>3</math> occurs every <math>3</math> terms, and starting out on the third term of the denominator, a factor of <math>3</math> appears also every <math>3</math> terms. Thus, the factors of <math>3</math> on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form <math>1 \cdot 2^a</math>. Since there will be no odd factors in the denominators, all the denominators will be in the form <math>2^a</math> where <math>a</math> is the number of factors of <math>2</math> in <math>(2009 \cdot 2)! = 4018!</math>. This is simply <math>\sum_{n=1}^{11} \left \lfloor{\frac{4018}{2^n}}\right \rfloor = 4010</math>. Therefore, our answer is <math>\boxed{401}</math>. |
+ | |||
+ | ==Solution 4== | ||
+ | Using the initial steps from Solution 1, <math>S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}</math>. Clearly <math>b = 1</math> as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find <math>a</math>. Note that <math>2^a</math> would be the largest denominator in all the summands, so when they are summed it is the common denominator. | ||
+ | |||
+ | Taking the p-adic valuations of each term, the powers of 2 in the denominator for <math>i</math> is <math>(2i) - v_2(\binom{2i}{i}).</math> We can use Kummers theorem to see that <math>v_2(\binom{2i}{i})</math> is the number of digits carried over when <math>i</math> is added to <math>i</math> in base <math>2</math>. This is simply the number of <math>1</math>'s in the binary representation of <math>i</math>. | ||
+ | |||
+ | Looking at the binary representations of some of the larger <math>i</math> we see <math>2009 = 11111011001_2</math> having eight <math>1</math>'s. So the power of two is <math>2 \cdot 2009 - 8 = 4010</math>. Experimenting with <math>2008, 2007, 2006</math> we see that the power of two are all <math>< 4010</math>, and under <math>2005</math> the power of two <math>2i - v_2(\binom{2i}{i}) < 2i < 4010</math>. Therefore <math>a = 4010</math> and <math>\frac{ab}{10} = \boxed{401}.</math> | ||
+ | |||
+ | ~Aaryabhatta1 | ||
+ | |||
+ | |||
+ | |||
+ | == Video Solution == | ||
+ | |||
+ | https://youtu.be/ppJkqLd3VNI?si=GX3UXkQ5s-5L6AzO | ||
+ | |||
+ | ~MathProblemSolvingSkills.com | ||
+ | |||
== See Also == | == See Also == |
Latest revision as of 16:23, 30 March 2024
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 you 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
.
Solution 3 (Divisibility)
We can logically deduce that the value will be 1. Listing out the first few values of odd and even integers, we have:
and
. Obviously, none of the factors of
in the denominator will cancel out, since the numerator is odd. Starting on the second term of the numerator, a factor of
occurs every
terms, and starting out on the third term of the denominator, a factor of
appears also every
terms. Thus, the factors of
on the denominator will always cancel out. We can apply the same logic for every other odd factor, so once terms all cancel out, the denominator of the final expression will be in the form
. Since there will be no odd factors in the denominators, all the denominators will be in the form
where
is the number of factors of
in
. This is simply
. Therefore, our answer is
.
Solution 4
Using the initial steps from Solution 1, . Clearly
as in all the summands there are no non-power of 2 factors in the denominator. So we seek to find
. Note that
would be the largest denominator in all the summands, so when they are summed it is the common denominator.
Taking the p-adic valuations of each term, the powers of 2 in the denominator for is
We can use Kummers theorem to see that
is the number of digits carried over when
is added to
in base
. This is simply the number of
's in the binary representation of
.
Looking at the binary representations of some of the larger we see
having eight
's. So the power of two is
. Experimenting with
we see that the power of two are all
, and under
the power of two
. Therefore
and
~Aaryabhatta1
Video Solution
https://youtu.be/ppJkqLd3VNI?si=GX3UXkQ5s-5L6AzO
~MathProblemSolvingSkills.com
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.