jmc

$$\begin{gathered} {\log }_8{a}_1+{\log }_8{a}_2+\dots +{\log }_8{a}_{12}={\log }_{8}a+{\log }_8(ar)+\dots +{\log }_8(a{r}^{11}) \\ ={\log }_8(a\cdot ar\cdot a{r}^2\cdot \cdots \cdot a{r}^{11})={\log }_8(a^{12}{r}^{66}) \end{gathered}$$ 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 $1001=91\ast 11$. 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 $\boxed{46}$. For the step above, you may also simply do $1001/11+1=91+1=92$ to find how many multiples of $11$ there are in between $11$ and $1001$. Then, divide $92/2$ = $\boxed{46}$ to find only the odd solutions.