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

Latest revision as of 14:15, 23 August 2009

Redirect to:

@@ Line 1: / Line 1: @@
-== Problem ==
+#REDIRECT [[2006 AIME I Problems/Problem 11]]
-A sequence is defined as follows <math> a_1=a_2=a_3=1, </math> and, for all positive integers <math> n, a_{n+3}=a_{n+2}+a_{n+1}+a_n. </math> Given that <math> a_{28}=6090307, a_{29}=11201821, </math> and <math> a_{30}=20603361, </math> find the remainder when <math> \displaystyle \sum^{28}_{k=1} a_k </math> is divided by 1000.
-== Solution ==
-== See also ==
-*[[2006 AIME II Problems]]
-[[Category:Intermediate Algebra Problems]]