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 $\sum _{r=1}^{12}{\log }_8{a}_r=2010$ is
(correct answer + 1, wrong answer - 0.25)

• A. $40$
• B. $42$
• C. $44$
• D. $46$

Answer 1

The correct option is D $46$

Given: $\sum _{r=1}^{12}{\log }_8{a}_r=2010$
$\Rightarrow {\log }_8{a}_1+{\log }_8{a}_2+{\log }_8{a}_3+\cdots +{\log }_8{a}_{12}=2010$
$\Rightarrow {\log }_8[a_1{a}_2{a}_3\cdots a_{12}]=2010$
$\Rightarrow a.ar.a{r}^2.a{r}^3\cdots a{r}^{11}=8^{2010}$
$\Rightarrow a^{12}{r}^{66}=2^{6030}$
$\Rightarrow (a^2{r}^{11}{)}^6=(2^{1005}{)}^6$
$\Rightarrow a^2{r}^{11}=2^{1005}$

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.
$\therefore \beta =1,3,5,7,...,91,$ which are $46.$