Problem 1397

1. Given the sequence $a_1, a_2, \dots, a_{13}$ is a geometric sequence with $a_1 = a$ and common ratio $r$, we can express the terms of the sequence as:
$$a_1 = a, \quad a_2 = ar, \quad a_3 = ar^2, \quad \ldots, \quad a_{13} = ar^{12}$$

2. The logarithmic sum given in the problem is:
$$\log_{2015} a_1 + \log_{2015} a_2 + \dots + \log_{2015} a_{13} = 2015$$

3. Using the properties of logarithms, we can combine the terms:
$$\log_{2015} a_1 + \log_{2015} a_2 + \dots + \log_{2015} a_{13} = \log_{2015} (a \cdot ar \cdot ar^2 \cdot \ldots \cdot ar^{12})$$

4. Simplifying the product inside the logarithm:
$$a \cdot ar \cdot ar^2 \cdot \ldots \cdot ar^{12} = a^{13} \cdot r^{0+1+2+\ldots+12} = a^{13} \cdot r^{78}$$

5. Therefore, the equation becomes:
$$\log_{2015} (a^{13} \cdot r^{78}) = 2015$$

6. Using the property of logarithms $\log_b (x \cdot y) = \log_b x + \log_b y$, we get:
$$\log_{2015} (a^{13} \cdot r^{78}) = \log_{2015} a^{13} + \log_{2015} r^{78} = 13 \log_{2015} a + 78 \log_{2015} r$$

7. Given that this sum equals 2015:
$$13 \log_{2015} a + 78 \log_{2015} r = 2015$$

8. Removing the logarithm by exponentiating both sides with base 2015, we get:
$$a^{13} \cdot r^{78} = 2015^{2015}$$

9. Taking the 13th root of both sides:
$$(a^{13} \cdot r^{78})^{1/13} = (2015^{2015})^{1/13}$$
$$a \cdot r^6 = 2015^{155}$$

10. Since $2015 = 5 \cdot 13 \cdot 31$, we can express $2015^{155}$ as:
$$2015^{155} = (5 \cdot 13 \cdot 31)^{155} = 5^{155} \cdot 13^{155} \cdot 31^{155}$$

11. We need to find the number of possible ordered pairs $(a, r)$ such that $a \cdot r^6 = 5^{155} \cdot 13^{155} \cdot 31^{155}$.

12. Each prime factor of $r$ can take on 26 different degrees, ranging from 0 to 25, because $6 \cdot 25 = 150 < 155$. Since $r$ has 3 prime factors (5, 13, and 31), the number of possible values for $r$ is:
$$26^3$$

13. For each value of $r$, $a$ is uniquely determined by the equation $a = \frac{2015^{155}}{r^6}$.

14. Therefore, the number of possible ordered pairs $(a, r)$ is:
$$\boxed{26^3}$$