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Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 6"

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A <math>\frac 1p</math> -array is a structured, infinite, collection of numbers. For example, a <math>\frac 13</math> -array is constructed as follows:  
 
A <math>\frac 1p</math> -array is a structured, infinite, collection of numbers. For example, a <math>\frac 13</math> -array is constructed as follows:  
 
   
 
   
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<center><math>\begin{align*}
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<math>\begin{align*}
 
1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\
 
1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\
 
\frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\
 
\frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\
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\frac 1{216} \qquad &\cdots\\
 
\frac 1{216} \qquad &\cdots\\
 
&\ddots
 
&\ddots
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\end{align*}</math></center>
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\end{align*}</math>
 
   
 
   
 
In general, the first entry of each row is <math>\frac{1}{2p}</math> times the first entry of the previous row. Then, each succeeding term in a row is <math>\frac 1p</math> times the previous term in the same row. If the sum of all the terms in a <math>\frac{1}{2008}</math> -array can be written in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find the remainder when <math>m+n</math> is divided by <math>2008</math>.
 
In general, the first entry of each row is <math>\frac{1}{2p}</math> times the first entry of the previous row. Then, each succeeding term in a row is <math>\frac 1p</math> times the previous term in the same row. If the sum of all the terms in a <math>\frac{1}{2008}</math> -array can be written in the form <math>\frac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers, find the remainder when <math>m+n</math> is divided by <math>2008</math>.

Revision as of 18:13, 29 January 2020

Problem 6

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

$\begin{align*} 1 \qquad \frac 13\,\ \qquad \frac 19\,\ \qquad \frac 1{27} \qquad &\cdots\\ \frac 16 \qquad \frac 1{18}\,\ \qquad \frac{1}{54} \qquad &\cdots\\ \frac 1{36} \qquad \frac 1{108} \qquad &\cdots\\ \frac 1{216} \qquad &\cdots\\ &\ddots \end{align*}$ (Error compiling LaTeX. Unknown error_msg)

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 $r$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*}$ (Error compiling LaTeX. Unknown error_msg)

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 $2008$.

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
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