Difference between revisions of "2006 AIME A Problems/Problem 15"

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== Problem ==
 
== Problem ==
 
Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math>
 
Given that a sequence satisfies <math> x_0=0 </math> and <math> |x_k|=|x_{k-1}+3| </math> for all integers <math> k\ge 1, </math> find the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}|. </math>
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{{Delete|The correct page that is linked to contains this solution and others}}
  
 
== Solution ==
 
== Solution ==

Revision as of 12:53, 23 May 2009

Problem

Given that a sequence satisfies $x_0=0$ and $|x_k|=|x_{k-1}+3|$ for all integers $k\ge 1,$ find the minimum possible value of $|x_1+x_2+\cdots+x_{2006}|.$

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Solution

If you take the recursion formula and square both sides, you get $x_k^2=x_{k-1}^2+6x_{k-1}+9$, or $x_k^2-x_{k-1}^2=6x_{k-1}+9$. If you take the sum of this from 1 to 2007, you see that

$\begin{eqnarray*}\sum_{k=1}^{2007}x_k^2-x_{k-1}^2=\sum_{k=1}^{2007}6x_{k-1}+9\\

x_{2007}^2-x_0^2=6(x_0+x_1+x_2+\cdots +x_{2006})+9*2007\\ x_{2007}^2=6(S)+9*2007\\

S=\frac{x_{2007}^2-9*2007}{6}\end{eqnarray*}$ (Error compiling LaTeX. Unknown error_msg)

where $S=x_0+x_1+x_2+\cdots +x_{2006}=x_1+x_2+\cdots +x_{2006}$.

Note that $x_{2007}$ is a multiple of 3, and the closest square multiple of 3 to 2*2007 is $(3*45)^2=9*2025$. Therefore the smallest value of the absolute value of $S$ is

$|S|=\left|\frac{9*2025-9*2007}{6}\right|=\frac{9*18}{6}=\boxed{027}$.

See also

2006 AIME II (ProblemsAnswer KeyResources)
Preceded by
Problem 14
Followed by
Final Problem
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions