2009 AIME II Problems/Problem 7

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

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 2s 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.

2009 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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2009 AIME II Problems/Problem 7

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