Difference between revisions of "2006 AIME II Problems/Problem 11"
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== Problem == | == Problem == | ||
| − | 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>\sum^{28}_{k=1} a_k </math> is divided by 1000. | + | 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>\sum^{28}_{k=1} a_k </math> is divided by 1000. |
== Solution == | == Solution == | ||
| − | Define the sum as <math> | + | Define the sum as <math>s</math>. Since <math>a_n\ = a_{n + 3} - a_{n + 2} - a_{n + 1} </math>, the sum will be: |
| − | :<math> | + | :<math>s = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}</math> |
| − | :<math> | + | :<math>s = (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 | + | The first two groups [[telescope]]. The third resembles <math>s</math>. |
| − | :<math> | + | :<math>s\ = a_1 - a_3 + a_{28} + a_{30} - s</math> |
| − | :<math> | + | :<math>2s\ = a_{28} + a_{30}</math> |
| − | :<math> | + | :<math>s\ = \frac{a_{28} + a_{30}}{2}</math> |
| − | <math>a_{28}</math> and <math>a_{30}</math> are both given; the last four digits of | + | <math>a_{28}</math> and <math>a_{30}</math> are both given; the last four digits of their sum is <math>3668</math>, and half of that is <math>1834</math>. Therefore, the answer is <math>834</math>. |
== See also == | == See also == | ||
Revision as of 13:54, 25 November 2007
Problem
A sequence is defined as follows 
and, for all positive integers 
Given that 
and 
find the remainder when 
is divided by 1000.
Solution
Define the sum as 
. Since 
, the sum will be:
The first two groups telescope. The third resembles .
and 
are both given; the last four digits of their sum is 
, and half of that is 
. Therefore, the answer is 
.
See also
| 2006 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
