2006 AIME I Problems/Problem 15: Difference between revisions
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== Solution == | == Solution == | ||
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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 
and 
for all integers 
find the minimum possible value of 
Solution
Since we may phrase the sequence as 
, 
, 
, and 
, the sun of these 4 terms is 0. Since 
, 
.
Therefore, the minimum possible value of 
See also
| 2006 AIME I (Problems • Answer Key • Resources) | ||
| 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 | ||
