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

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

$= \log_8(a\cdot ar\cdot ar^2\cdot \cdots \cdot ar^{11}) = \log_8 (a^{12}r^{66})$

So our question is equivalent to solving $\log_8 (a^{12}r^{66})=2006$ for $a, r$ positive integers.

$a^{12}r^{66}=8^{2006} = (2^3)^{2006} = (2^6)^{1003}$ so

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

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 themselves powers of 2. Now, let $a=2^x$ and $r=2^y$ :

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

$2^{2x}\cdot 2^{11y}=2^{1003}$

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

$2x+11y=1003$

$11y=1003-2x$

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

For $y$ to be an integer, the numerator must be divisible by 11. This occurs when $x=1$ because $1001=91*11$ . Because only even integers are being subtracted from 1003, the numerator never equals an even multiple of 11. Therefore, the numerator takes on the value of every odd multiple of 11 from 11 to 1001. Since the odd multiples are separated by a distance of 22, the number of ordered pairs that work is $1 + \frac{1001-11}{22}=1 + \frac{990}{22}=46$ . (We must add 1 because both endpoints are being included.) So the answer is $46$ .

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2006 AIME I Problems/Problem 9

Problem

Solution

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