Difference between revisions of "2009 AIME II Problems/Problem 7"

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Problem

Define $n!!$ to be $n(n-2)(n-4)\cdots 3\cdot 1$ for $n$ odd and $n(n-2)(n-4)\cdots 4\cdot 2$ for $n$ even. When $\sum_{i=1}^{2009} \frac{(2i-1)!!}{(2i)!!}$ is expressed as a fraction in lowest terms, its denominator is $2^ab$ with $b$ odd. Find $\dfrac{ab}{10}$ .

Solution 1

First, note that $(2n)!! = 2^n \cdot n!$ , and that $(2n)!! \cdot (2n-1)!! = (2n)!$ .

We can now take the fraction $\dfrac{(2i-1)!!}{(2i)!!}$ and multiply both the numerator and the denominator by $(2i)!!$ . We get that this fraction is equal to $\dfrac{(2i)!}{(2i)!!^2} = \dfrac{(2i)!}{2^{2i}(i!)^2}$ .

Now we can recognize that $\dfrac{(2i)!}{(i!)^2}$ is simply ${2i \choose i}$ , hence this fraction is $\dfrac{{2i\choose i}}{2^{2i}}$ , and our sum turns into $S=\sum_{i=1}^{2009} \dfrac{{2i\choose i}}{2^{2i}}$ .

Let $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ . Obviously $c$ is an integer, and $S$ can be written as $\dfrac{c}{2^{2\cdot 2009}}$ . Hence if $S$ is expressed as a fraction in lowest terms, its denominator will be of the form $2^a$ for some $a\leq 2\cdot 2009$ .

In other words, we just showed that $b=1$ . To determine $a$ , we need to determine the largest power of $2$ that divides $c$ .

Let $p(i)$ be the largest $x$ such that $2^x$ that divides $i$ .

We can now return to the observation that $(2i)! = (2i)!! \cdot (2i-1)!! = 2^i \cdot i! \cdot (2i-1)!!$ . Together with the obvious fact that $(2i-1)!!$ is odd, we get that $p((2i)!)=p(i!)+i$ .

It immediately follows that $p\left( {2i\choose i} \right) = p((2i)!) - 2p(i!) = i - p(i!)$ , and hence $p\left( {2i\choose i} \cdot 2^{2\cdot 2009 - 2i} \right) = 2\cdot 2009 - i - p(i!)$ .

Obviously, for $i\in\{1,2,\dots,2009\}$ the function $f(i)=2\cdot 2009 - i - p(i!)$ is is a strictly decreasing function. Therefore $p(c) = p\left( {2\cdot 2009\choose 2009} \right) = 2009 - p(2009!)$ .

We can now compute $p(2009!) = \sum_{k=1}^{\infty} \left\lfloor \dfrac{2009}{2^k} \right\rfloor = 1004 + 502 + \cdots + 3 + 1 = 2001$ . Hence $p(c)=2009-2001=8$ .

And thus we have $a=2\cdot 2009 - p(c) = 4010$ , and the answer is $\dfrac{ab}{10} = \dfrac{4010\cdot 1}{10} = \boxed{401}$ .

Additionally, once you count the number of factors of $2$ in the summation, one can consider the fact that, since $b$ must be odd, it has to take on a value of $1,3,5,7,$ or $9$ (Because the number of $2$ s in the summation is clearly greater than $1000$ , dividing by $10$ will yield a number greater than $100$ , and multiplying this number by any odd number greater than $9$ will yield an answer $>999$ , which cannot happen on the AIME.) Once you calculate the value of $4010$ , and divide by $10$ , $b$ must be equal to $1$ , as any other value of $b$ will result in an answer $>999$ . This gives $\boxed{401}$ as the answer.

Just a small note. It's important to note the properties of the $v_p$ function, which is what Solution 1 is using but denoting it as $p (...)$ . We want to calculate $v_2 \left( \sum ^{2009} _{i=1} \dbinom{2i}{i} \cdot 2^{2 \cdot 2009 - 2i} \right)$ as the final step. We know that one property of $v_p$ is that $v_p (a + b) \geq \min \left( v_p(a), v_p (b) \right)$ . Therefore, we have that $v_2 \left( \sum ^{2009} _{i=1} \dbinom{2i}{i} \cdot 2^{2 \cdot 2009 - 2i} \right) \geq \min \left( 2 \cdot 2009 -1, 2 \cdot 2009 -2 - 1, ... , 2 \cdot 2009 - 2008 - v_2 (2008!), 2 \cdot 2009 - 2009 - v_2 (2009!) \right)$ . Thus, we see by similar calculations as in Solution 1, that $v_2 (c) \geq 8$ . From which the conclusion follows.

- (OmicronGamma)

Solution 2

Using the steps of the previous solution we get $c = \sum_{i=1}^{2009} {2i\choose i} \cdot 2^{2\cdot 2009 - 2i}$ and if you do the small cases(like $1, 2, 3, 4, 5, 6$ ) you realize that you can "thin-slice" the problem and simply look at the cases where $i=2009, 2008$ (they're nearly identical in nature but one has $4$ with it) since $\dbinom{2i}{I}$ hardly contains any powers of $2$ 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 $2$ in $\dbinom{4018}{2009}$ and $\dbinom{4016}{2008}$ and you get the minimum power of $2$ in either expression is $8$ so the answer is $\frac{4010}{10} \implies \boxed{401}$ since it would violate the rules of the AIME and the small cases if $b>1$ .

Solution 3

Realize that the most powers of two comes in the last term, and calculate the number of powers of 2 in it by taking the floor of 4018 when divided by 2, by 4, etc. This result comes out to $4010$ , and so the only possible odd number is $1$ as it would otherwise violate AIME rules.

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 {{AIME box|year=2009|n=II|num-b=6|num-a=8}}
-[[Category: Intermediate Number Theory problems]]
+[[Category: Intermediate Number Theory Problems]]
 {{MAA Notice}}

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