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

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== Solution ==
 
== Solution ==
 
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
 
<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>
+
     \log_8 a+log_8 (ar)+\ldots+\log_8 (ar^{11})</math>
 +
 
 
We must now use the rules of logarithms:
 
We must now use the rules of logarithms:
 +
 
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
 
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}=
−
     \log_8 (a*ar*ar^2*\cdots*ar^{11}*ar^{12}</math>
+
     \log_8 (a*ar*ar^2*\cdots*ar^{11})</math>
 +
 
 +
<math>\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 (a^{12}r^{66})</math>
 +
 
 +
<math>\log_8 (a^{12}r^{66})=2006</math>
 +
 
 +
<math>a^{12}r^{66}=8^{2006}</math>
  
 
== See also ==
 
== See also ==

Revision as of 13:52, 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

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+log_8 (ar)+\ldots+\log_8 (ar^{11})$

We must now use the rules of logarithms:

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 (a*ar*ar^2*\cdots*ar^{11})$

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 (a^{12}r^{66})$

$\log_8 (a^{12}r^{66})=2006$

$a^{12}r^{66}=8^{2006}$

See also

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