

(4 intermediate revisions by 2 users not shown) 
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 [[integer]]s <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 ==
 
−  Define the sum as <math>x</math>. Notice that <math>a_n\ = a_{n + 3}  a_{n + 2}  a_{n + 1} </math>, so the sum will be:
 
−  :<math>x = (a_4  a_3  a_2) + (a_5  a_4  a_3) + \ldots (a_{30}  a_{29}  a_{28}) + a_{28}</math>
 
−  :<math>x = (a_4+ a_5 \ldots a_{30})  (a_3 + a_4 + \ldots a_{29})  (a_2 + a_3 + \ldots a_{28}) + a_{28} + (a_1  a_1)</math>
 
−   
−  The first two groupings almost completely cancel. The third resembles <math>x</math>.
 
−   
−  :<math>x\ = a_1  a_3 + a_{28} + a_{30}  x</math>
 
−  :<math>2x\ = a_{28} + a_{30}</math>
 
−  :<math>x\ = \frac{a_{28} + a_{30}}{2}</math>
 
−   
−  <math>a_{28}</math> and <math>a_{30}</math> are both given; the last four digits of the sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>834</math>.
 
−   
−  == See also ==
 
−  *[[2006 AIME II Problems]]
 
−   
−  {{AIME boxyear=2006n=IInumb=10numa=12}}
 
−   
−  [[Category:Intermediate Algebra Problems]]
 