Question

Let $a_1,a_2,a_3,\dots$ be a G.P. with $a_1=a$ and common ratio $r,$ where $a$ and $r$ positive integers, then the number of ordered pairs $(a,r)$ such that $\sum _{k=1}^{12}{\log }_8{a}_k=2010$ is

Answer 1

Given: $\sum _{k=1}^{12}{\log }_8{a}_k=2010$
$\Rightarrow {\log }_8{a}_1+{\log }_8{a}_2+{\log }_8{a}_3+\cdots +{\log }_8{a}_{12}=2010\Rightarrow {\log }_8[a_1{a}_2{a}_3\cdots a_{12}]=2010\Rightarrow a.ar.a{r}^2.a{r}^3\cdots a{r}^{11}=8^{2010}\Rightarrow a^{12}{r}^{66}=2^{6030}\Rightarrow a^2{r}^{11}=2^{1005}$

Let $a=2^{\alpha },r=2^{\beta },$ where $\alpha ,\beta$ are non negative integers
$\Rightarrow 2\alpha +11\beta =1005$
If $\alpha =0,$ then
$\beta =\left[\frac{1005}{11}\right]=91$
$\Rightarrow \beta \le 91$
Also, $11\beta =1005-2\alpha$
Which is odd, so $\beta$ is also odd, so
$\beta =1,3,5,7,...,91$
Therefore, the number of terms
$1+(n-1)\times 2=91\Rightarrow n=46$

Hence, there are $46$ pairs of $(a,r)$ possible.