Let $\Gamma$ be the maximum possible value of $a+3b+9c$ among all triples $(a,b,c)$ of positive real numbers such that
\[ \log_{30}(a+b+c) = \log_{8}(3a) = \log_{27} (3b) = \log_{125} (3c) .\]
If $\Gamma = \frac{p}{q}$ where $p$ and $q$ are relatively prime positive integers, then find $p+q$.
1. We start by setting all the logarithmic expressions equal to a common variable \( k \):
   \[
   \log_{30}(a+b+c) = k, \quad \log_{8}(3a) = k, \quad \log_{27}(3b) = k, \quad \log_{125}(3c) = k
   \]
   
2. Converting these logarithmic equations to exponential form, we get:
   \[
   30^k = a + b + c
   \]
   \[
   8^k = 3a \implies 2^{3k} = 3a \implies a = \frac{2^{3k}}{3}
   \]
   \[
   27^k = 3b \implies 3^{3k} = 3b \implies b = 3^{3k-1}
   \]
   \[
   125^k = 3c \implies 5^{3k} = 3c \implies c = \frac{5^{3k}}{3}
   \]

3. We now multiply the expressions for \(a\), \(b\), and \(c\):
   \[
   a \cdot b \cdot c = \left(\frac{2^{3k}}{3}\right) \cdot 3^{3k-1} \cdot \left(\frac{5^{3k}}{3}\right) = \frac{2^{3k} \cdot 3^{3k-1} \cdot 5^{3k}}{9}
   \]

4. Simplifying the product:
   \[
   a \cdot b \cdot c = \frac{2^{3k} \cdot 3^{3k-1} \cdot 5^{3k}}{9} = \frac{(2 \cdot 3 \cdot 5)^{3k}}{9 \cdot 3} = \frac{30^{3k}}{27}
   \]

5. Using the equation \(30^k = a + b + c\), we substitute \(a + b + c\) into the product:
   \[
   (a + b + c)^3 = 30^{3k}
   \]
   \[
   (a + b + c)^3 = 27 \cdot a \cdot b \cdot c
   \]

6. From the AM-GM inequality, we know:
   \[
   \frac{a + b + c}{3} \geq \sqrt[3]{abc}
   \]
   Equality holds when \(a = b = c\). Therefore, we set \(a = b = c\).

7. Substituting \(a = b = c\) into the logarithmic equations, we get:
   \[
   2^{3k} = 3a, \quad 3^{3k} = 3b, \quad 5^{3k} = 3c
   \]
   Since \(a = b = c\), we have:
   \[
   2^{3k} = 3^{3k} = 5^{3k}
   \]
   This implies \(k = 0\).

8. Substituting \(k = 0\) back into the equations, we get:
   \[
   a = b = c = \frac{1}{3}
   \]

9. Finally, we calculate \(a + 3b + 9c\):
   \[
   a + 3b + 9c = \frac{1}{3} + 3 \left(\frac{1}{3}\right) + 9 \left(\frac{1}{3}\right) = \frac{1}{3} + 1 + 3 = \frac{13}{3}
   \]

10. Since \(\Gamma = \frac{13}{3}\), and \(p = 13\) and \(q = 3\) are relatively prime, we find:
   \[
   p + q = 13 + 3 = 16
   \]

The final answer is \(\boxed{16}\).