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| − | == 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.
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| − | == Solution ==
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| − | 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:
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| − | :<math>x = (a_4 - a_3 - a_2) + (a_5 - a_4 - a_3) + \ldots (a_{30} - a_{29} - a_{28}) + a_{28}</math>
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| − | :<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>
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| − | The first two groupings almost completely cancel. The third resembles <math>x</math>.
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| − | :<math>x\ = a_1 - a_3 + a_{28} + a_{30} - x</math>
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| − | :<math>2x\ = a_{28} + a_{30}</math>
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| − | :<math>x\ = \frac{a_{28} + a_{30}}{2}</math>
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| − | <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>.
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| − | == See also ==
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| − | *[[2006 AIME II Problems]]
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| − | {{AIME box|year=2006|n=II|num-b=10|num-a=12}}
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| − | [[Category:Intermediate Algebra Problems]]
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