AoPSWiki
Want to learn how to tackle those tough MATHCOUNTS and AMC counting and probability problems? Check out Art of Problem Solving's Introduction to Counting & Probability by David Patrick.
Personal tools

Mock AIME 1 2007-2008 Problems/Problem 6

From AoPSWiki

Problem 6

A \frac 1p -array is a structured, infinite, collection of numbers. For example, a \frac 13 -array is constructed as follows:

\begin{align*}1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\\frac 16 \qquad \frac 1{18}\,\ \...

In general, the first entry of each row is \frac{1}{2p} times the first entry of the previous row. Then, each succeeding term in a row is \frac 1p times the previous term in the same row. If the sum of all the terms in a \frac{1}{2008} -array can be written in the form \frac mn, where m and n are relatively prime positive integers, find the remainder when m+n is divided by 2008.

Solution

Note that the value in the rth row and the cth 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(\su...

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}{2 \cdot 2008} \cdo.... 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.

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 5
Followed by
Problem 7
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Looking for a challenging geometry text? Preparing for MATHCOUNTS or the AMC exams? Check out Art of Problem Solving's Introduction to Geometry by Richard Rusczyk.
© Copyright 2008 AoPS Incorporated. All Rights Reserved. • FoundationPrivacyContact Us