Mock AIME 1 2007-2008 Problems/Problem 6

Problem 6

[problem to be filled in]

Solution

Consider the product of the geometric series $\left(1+\frac{1}{2 \cdot 2008} + \frac{1}{4 \cdot 2008^2} \cdots\right)\left(1 + \frac{1}{2008} + \frac{1}{2 \cdot 2008} \cdots \right)$ .

By the geometric series formula, the first series evaluates to be $\frac{1}{1 - \frac{1}{2 \cdot 2008}} = \frac{2 \cdot 2008}{2 \cdot 2008 - 1}$ . The second series evaluates to be $\frac{1}{1 - \frac{1}{2008}} = \frac{2008}{2008 - 1}$ . Their product is $\frac{2008 \cdot 4016}{(2008-1)(2\cdot 2008 - 1)}$ . Taking the denominator, we have $(2008-1)(2\cdot 2008 - 1) \equiv (-1)(-1) \equiv 1 \pmod{2008}$ (or consider FOILing). The answer is $\boxed{001}$ .

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Mock AIME 1 2007-2008 (Problems, Source)
Preceded by Problem 5	Followed by Problem 7
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Mock AIME 1 2007-2008 Problems/Problem 6

Problem 6

Solution

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