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2006 AIME I Problems/Problem 9: Difference between revisions

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== Solution ==
== Solution ==
 
<math>\log(a1)+\log(a2)+\log(a3)+\cdots+\log(a12)=</math>
 
 


== See also ==
== See also ==

Revision as of 13:41, 3 August 2006

Problem

The sequence $a_1, a_2, \ldots$ is geometric with $a_1=a$ and common ratio $r,$ where $a$ and $r$ are positive integers. Given that $\log_8 a_1+\log_8 a_2+\cdots+\log_8 a_{12} = 2006,$ find the number of possible ordered pairs $(a,r).$



Solution

$\log(a1)+\log(a2)+\log(a3)+\cdots+\log(a12)=$

See also

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