We define $\lfloor x \rfloor$ as the largest integer less than or equal to $x$. What is
\[ \left\lfloor \frac{5^{2017015}}{5^{2015}+7} \right\rfloor \mod 1000?\]
1. We start by simplifying the expression \(\left\lfloor \frac{5^{2017015}}{5^{2015} + 7} \right\rfloor\). Notice that \(5^{2017015}\) is a very large number, and \(5^{2015} + 7\) is relatively small in comparison. We can use the method of successive approximations to simplify the division.

2. First, we approximate:
   \[
   \frac{5^{2017015}}{5^{2015} + 7} \approx \frac{5^{2017015}}{5^{2015}} = 5^{2017015 - 2015} = 5^{2015000}
   \]
   This approximation is very close, but we need to account for the small term \(7\) in the denominator.

3. We use the method of long division to refine our approximation:
   \[
   \frac{5^{2017015}}{5^{2015} + 7} = 5^{2015000} - \frac{7 \cdot 5^{2015000}}{5^{2015} + 7}
   \]
   The term \(\frac{7 \cdot 5^{2015000}}{5^{2015} + 7}\) is very small compared to \(5^{2015000}\), so we can continue this process iteratively:
   \[
   \frac{7 \cdot 5^{2015000}}{5^{2015} + 7} \approx 7 \cdot 5^{2015000 - 2015} = 7 \cdot 5^{2014985}
   \]
   Thus,
   \[
   \left\lfloor \frac{5^{2017015}}{5^{2015} + 7} \right\rfloor = 5^{2015000} - 7 \cdot 5^{2014985} + \frac{7^2 \cdot 5^{2014985}}{5^{2015} + 7}
   \]
   Continuing this process, we get:
   \[
   \left\lfloor \frac{5^{2017015}}{5^{2015} + 7} \right\rfloor = 5^{2015000} - 7 \cdot 5^{2014985} + 7^2 \cdot 5^{2014970} - \cdots
   \]

4. To find the value modulo 1000, we need to consider the terms modulo 1000. Notice that \(5^{2015000}\) and higher powers of 5 will be multiples of 1000, so they contribute 0 modulo 1000.

5. Therefore, we need to consider the sum of the series modulo 1000:
   \[
   \left\lfloor \frac{5^{2017015}}{5^{2015} + 7} \right\rfloor \equiv -1 + \sum_{k=0}^{1000} (-7)^{1000-k} \cdot 5^{2015k} \pmod{1000}
   \]

6. We simplify the sum modulo 1000. Notice that \(5^{2015k}\) for \(k \geq 3\) will be multiples of 1000, so we only need to consider the first few terms:
   \[
   -1 + \sum_{k=0}^{2} (-7)^{1000-k} \cdot 5^{2015k} \pmod{1000}
   \]

7. Since \(7^{1000} \equiv 1 \pmod{1000}\) (by Euler's theorem), we have:
   \[
   -1 + 1 \cdot 5^0 + (-7) \cdot 5^{2015} + 7^2 \cdot 5^{4030} \equiv -1 + 1 - 7 \cdot 0 + 49 \cdot 0 \equiv 0 \pmod{1000}
   \]

The final answer is \(\boxed{0}\).