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Difference between revisions of "2006 AIME A Problems/Problem 11"

 
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== See also ==
 
== See also ==
 
*[[2006 AIME II Problems]]
 
*[[2006 AIME II Problems]]
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[[Category:Intermediate Algebra Problems]]

Revision as of 09:39, 16 July 2006

Problem

A sequence is defined as follows $a_1=a_2=a_3=1,$ and, for all positive integers $n, a_{n+3}=a_{n+2}+a_{n+1}+a_n.$ Given that $a_{28}=6090307, a_{29}=11201821,$ and $a_{30}=20603361,$ find the remainder when $\displaystyle \sum^{28}_{k=1} a_k$ is divided by 1000.

Solution

See also

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