Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 12"

Revision as of 20:13, 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, $(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1$

~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.

@@ Line 12: / Line 12: @@
 Therefore <math>a_k=A^{k+1}-1</math>, and <math>b_k=A^k</math>
-Then, $(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1
+Then, <math>(A)(b_k)-a_k=(A)(A^k)-(A^{k+1}-1)=A^{k+1}-A^{k+1}+1=1</math>
 ~Tomas Diaz. orders@tomasdiaz.com
 {{alternate solutions}}

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Difference between revisions of "Mock AIME 6 2006-2007 Problems/Problem 12"

Revision as of 20:13, 26 November 2023

Problem

Solution