2006 AIME I Problems/Problem 9

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

$\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 a+log_8 (a*r)+\ldots+\log_8 (a*r^{11})$ We must now use the rules of logarithms: $\log_8 a_1+\log_8 a_2+\ldots+\log_8 a_{12}= \log_8 (a*ar*ar^2*\cdots*ar^{11}*ar^{12}$

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2006 AIME I Problems/Problem 9

Problem

Solution

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