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2006 AIME I Problems/Problem 9

Revision as of 14:02, 3 August 2006 by Ninja glace (talk | contribs) (→‎Solution)

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

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

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

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

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

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

$a^{12}r^{66}=(2^3)^{2006}$

$a^{2}r^{11}=2^{1003}$

The product of $a^2$ and $r^{11}$ is a power of 2. Since both numbers have to be integers, this means that a and r are also powers of 2. Now, let $a=2^x$ and $r=2^y$:

$(2^x)^2*(2^y)^{11}=2^{1003}$

$2^{2x}*2^{11y}=2^{1003}$

$2^{2x+11y}=2^{1003}$

$2x+11y=1003$

$11y=1003-2x$

$y=\frac{1003-2x}{11}$

For y to be an integer, the numerator must be divisible by 2. This occurs when $x=1$ because $1001=91*11$.

See also

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