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Difference between revisions of "2006 AIME I Problems/Problem 9"

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== Solution ==
 
== Solution ==
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<math>\log(a1)+\log(a2)+\log(a3)+\ldots+\log(a12)=</math>
+
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_12=
 +
    \log_8 a+log_8 (a*r)+\ldots+\log_8 (a*r^11)</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:45, 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

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_12= \log_8 a+log_8 (a*r)+\ldots+\log_8 (a*r^11)$

See also

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