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(C) Given $\log_8(a_1 a_2 \dots a_{12}) = 2014$. Since $a_k = a_1 r^{k-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$. Thus,$\log_

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$46$

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(C) Given $\log_8(a_1 a_2 \dots a_{12}) = 2014$. Since $a_k = a_1 r^{k-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$. Thus,$\log_8(a_1^{12} r^{66}) = 2014 \Rightarrow a_1^{12} r^{66} = 8^{2014} = 2^{6042}$. Let $a_1 = 2^m$ and $r = 2^n$ where $m, n$ are integers. Then $(2^m)^{12} (2^n)^{66} = 2^{12m + 66n} = 2^{6042}$. This implies $12m + 66n = 6042$,which simplifies to $2m + 11n = 1007$. Since $r > 1$ and the progression is increasing,$n \ge 1$. Also $a_1 \ge 1$,so $m \ge 0$. $2m = 1007 - 11n$. For $m$ to be an integer,$1007 - 11n$ must be even,so $n$ must be odd. Also $m > 0$ $\Rightarrow 1007 - 11n > 0$ $\Rightarrow n Thus $n \in \{1, 3, 5, \dots, 91\}$. The number of such odd values is $\frac{91 - 1}{2} + 1 = 46$.

Given $a_1, a_2, a_3, \dots$ form an increasing geometric progression with common ratio $r$ such that $\log_8 a_1 + \log_8 a_2 + \dots + \log_8 a_{12} = 2014$,then the number of ordered pairs of integers $(a_1, r)$ is equal to

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