Difference between revisions of "Mock AIME 5 2005-2006 Problems/Problem 15"

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== Problem ==
 
== Problem ==
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<math>2006</math> colored beads are placed on a necklace (circular ring) such that each bead is adjacent to two others. The beads are labeled <math>a_0</math>, <math>a_1</math>, <math>\ldots</math>, <math>a_{2005}</math> around the circle in order. Two beads <math>a_i</math> and <math>a_j</math>, where <math>i</math> and <math>j</math> are non-negative integers, satisfy <math>a_i = a_j</math> if and only if the color of <math>a_i</math> is the same as the color of <math>a_j</math>. Given that there exists no non-negative integer <math>m < 2006</math> and positive integer <math>n < 1003</math> such that <math>a_m = a_{m-n} = a_{m+n}</math>, where all subscripts are taken <math>\pmod{2006}</math>, find the minimum number of different colors of beads on the necklace.
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== Solution ==
  
 
== Solution ==
 
== Solution ==

Latest revision as of 21:20, 8 October 2014

Problem

$2006$ colored beads are placed on a necklace (circular ring) such that each bead is adjacent to two others. The beads are labeled $a_0$, $a_1$, $\ldots$, $a_{2005}$ around the circle in order. Two beads $a_i$ and $a_j$, where $i$ and $j$ are non-negative integers, satisfy $a_i = a_j$ if and only if the color of $a_i$ is the same as the color of $a_j$. Given that there exists no non-negative integer $m < 2006$ and positive integer $n < 1003$ such that $a_m = a_{m-n} = a_{m+n}$, where all subscripts are taken $\pmod{2006}$, find the minimum number of different colors of beads on the necklace.

Solution

Solution

See also

Mock AIME 5 2005-2006 (Problems, Source)
Preceded by
Problem 14
Followed by
Problem 15
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