Mock AIME 1 2007-2008 Problems/Problem 6

what the sigma

Solution

Note that the value in the $r$ th row and the $c$ th column is given by $\left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right)$ . We wish to evaluate the summation over all $r,c$ , and so the summation will be, using the formula for an infinite geometric series: $\begin{align*}\sum_{r=1}^{\infty}\sum_{c=1}^{\infty} \left(\frac{1}{(2p)^r}\right)\left(\frac{1}{p^c}\right) &= \left(\sum_{r=1}^{\infty} \frac{1}{(2p)^r}\right)\left(\sum_{c=1}^{\infty} \frac{1}{p^c}\right)\\ &= \left(\frac{1}{1-\frac{1}{2p}}\right)\left(\frac{1}{1-\frac{1}{p}}\right)\\ &= \frac{2p^2}{(2p-1)(p-1)}\end{align*}$ Taking the denominator with $p=2008$ (indeed, the answer is independent of the value of $p$ ), we have $m+n \equiv 2008^2 + (2008-1)(2\cdot 2008 - 1) \equiv (-1)(-1) \equiv 1 \pmod{2008}$ (or consider FOILing). The answer is $\boxed{001}$ .

With less notation, the above solution is equivalent to considering the product of the geometric series $\left(1+\frac{1}{2 \cdot 2008} + \frac{1}{4 \cdot 2008^2} \cdots\right)\left(1 + \frac{1}{2008} + \frac{1}{2008^2} \cdots \right)$ . Note that when we expand this product, the terms cover all of the elements of the array.

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 $\frac{2008 \cdot 4016}{(2008-1)(2\cdot 2008 - 1)}$ , from which we find that $m+n$ leaves a residue of $1$ upon division by $2008$ .

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

Solution

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