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(C) Given the sum of logarithms: $\log_8(a_1 a_2 \dots a_{12}) = 2014$.Since $a_n = a_1 r^{n-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12

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$46$

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(C) Given the sum of logarithms: $\log_8(a_1 a_2 \dots a_{12}) = 2014$.Since $a_n = a_1 r^{n-1}$,the product is $a_1^{12} r^{0+1+2+\dots+11} = a_1^{12} r^{66}$.Thus,$\log_8(a_1^{12} r^{66}) = 2014$,which implies $a_1^{12} r^{66} = 8^{2014} = (2^3)^{2014} = 2^{6042}$.Let $a_1 = 2^m$ and $r = 2^n$ for integers $m, n$ (since the progression is increasing,$r > 1$,so $n \ge 1$).Substituting these into the equation: $(2^m)^{12} (2^n)^{66} = 2^{12m + 66n} = 2^{6042}$.Equating exponents: $12m + 66n = 6042$,which simplifies to $2m + 11n = 1007$.Solving for $m$: $m = \frac{1007 - 11n}{2}$.For $m$ to be an integer,$1007 - 11n$ must be even,meaning $n$ must be odd.Since $a_1$ and $r$ are integers and the progression is increasing,$n \ge 1$ and $m \ge 1$.$1007 - 11n \ge 2 \Rightarrow 11n \le 1005 \Rightarrow n \le 91.36$.Thus,$n$ can take odd values: $1, 3, 5, \dots, 91$.The number of such 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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