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

m
(Added a solution)
Line 3: Line 3:
  
 
== Solution ==
 
== Solution ==
{{solution}}
+
 
 +
Since we may phrase the sequence as <math> x_{4n}=0 </math>, <math> x_{4n+1}= \pm 3 </math>, <math> x_{4n+2}=0 </math>, and <math> x_{4n+3}= \mp 3 </math>, the sun of these 4 terms is 0. Since <math> {2006} \equiv {2} modulo {4} </math>, <math> |0 \pm 3|  = 3</math>.
 +
 
 +
Therefore, the minimum possible value of <math> |x_1+x_2+\cdots+x_{2006}| = 3 .</math>
 +
 
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}
 
{{AIME box|year=2006|n=I|num-b=14|after=Last Question}}

Revision as of 18:02, 10 October 2007

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 $|x_1+x_2+\cdots+x_{2006}|.$

Solution

Since we may phrase the sequence as $x_{4n}=0$, $x_{4n+1}= \pm 3$, $x_{4n+2}=0$, and $x_{4n+3}= \mp 3$, the sun of these 4 terms is 0. Since ${2006} \equiv {2} modulo {4}$, $|0 \pm 3|  = 3$.

Therefore, the minimum possible value of $|x_1+x_2+\cdots+x_{2006}| = 3 .$


See also

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