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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 12"
 
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Therefore <math>a_k=A^{k+1}-1</math>, and <math>b_k=A^k</math>
 
Therefore <math>a_k=A^{k+1}-1</math>, and <math>b_k=A^k</math>
  
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Then, $(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1
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Then, <math>(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1</math>
 +
 
 +
<math>\sum_{k=2}^{A} ((A)(b_k)-a_k)=\sum_{k=2}^{A}1=A-1</math>
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 +
Therefore, <math>S=2007-1=2006 \equiv 6\;(mod\;1000)</math>
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Reminder is <math>\boxed{6}</math>
  
 
~Tomas Diaz. orders@tomasdiaz.com
 
~Tomas Diaz. orders@tomasdiaz.com
  
 
{{alternate solutions}}
 
{{alternate solutions}}

Latest revision as of 20:17, 26 November 2023

Problem

Let $x_k$ be the largest positive rational solution $x$ to the equation $(2007-x)(x+2007^{-k})^k=1$ for all integers $k\ge 2$. For each $k$, let $x_k=\frac{a_k}{b_k}$, where $a_k$ and $b_k$ are relatively prime positive integers. If \[S=\sum_{k=2}^{2007} (2007b_k-a_k),\] what is the remainder when $S$ is divided by $1000$?

Solution

Let $A=2007$

Solving: $(A-x)(x+A^{-k})^k=1$ we note that the largest positive rational solution $x$ is given by:

$x_k=A-\frac{1}{A^k}=\frac{A^{k+1}-1}{A^k}=\frac{a_k}{b_k}$

Therefore $a_k=A^{k+1}-1$, and $b_k=A^k$

Then, $(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1$

$\sum_{k=2}^{A} ((A)(b_k)-a_k)=\sum_{k=2}^{A}1=A-1$

Therefore, $S=2007-1=2006 \equiv 6\;(mod\;1000)$

Reminder is $\boxed{6}$

~Tomas Diaz. orders@tomasdiaz.com

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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