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 $r=2^y$ :

$\begin{eqnarray*}(2^x)^2\cdot(2^y)^{11}&=&2^{1003}\\ 2^{2x}\cdot 2^{11y}&=&2^{1003}\\ 2x+11y&=&1003\\ y&=&\frac{1003-2x}{11} \end{eqnarray*}$

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 $\boxed{046}$ .

2006 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
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2006 AIME I Problems/Problem 9

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