2006 AIME I Problems/Problem 15

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 $|x_1+x_2+\cdots+x_{2006}| = 3 .$

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 14	Followed by Last Question
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2006 AIME I Problems/Problem 15

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