How many triples $(p,q,n)$ are there such that $1/p+2013/q = n/5$ where $p$, $q$ are prime numbers and $n$ is a positive integer?

$ 
\textbf{(A)}\ 7
\qquad\textbf{(B)}\ 6
\qquad\textbf{(C)}\ 5 
\qquad\textbf{(D)}\ 4
\qquad\textbf{(E)}\ 4
$
We start with the given equation:
\[
\frac{1}{p} + \frac{2013}{q} = \frac{n}{5}
\]
where \( p \) and \( q \) are prime numbers, and \( n \) is a positive integer. We can rewrite this equation as:
\[
\frac{2013p + q}{pq} = \frac{n}{5}
\]
Multiplying both sides by \( 5pq \) to clear the denominators, we get:
\[
5(2013p + q) = npq
\]
Rearranging terms, we have:
\[
npq = 5(2013p + q)
\]

### Case 1: \( p = 5 \)
Substitute \( p = 5 \) into the equation:
\[
n \cdot 5 \cdot q = 5(2013 \cdot 5 + q)
\]
Simplify:
\[
5nq = 5(10065 + q)
\]
Divide both sides by 5:
\[
nq = 10065 + q
\]
Rearrange to solve for \( n \):
\[
nq - q = 10065
\]
Factor out \( q \):
\[
q(n - 1) = 10065
\]
Since \( q \) is a prime number, we need to factorize 10065:
\[
10065 = 3 \times 5 \times 11 \times 61
\]
The possible prime values for \( q \) are 3, 11, and 61. For each of these values, we can find \( n \):
- If \( q = 3 \), then \( 3(n - 1) = 10065 \Rightarrow n - 1 = 3355 \Rightarrow n = 3356 \)
- If \( q = 11 \), then \( 11(n - 1) = 10065 \Rightarrow n - 1 = 915 \Rightarrow n = 916 \)
- If \( q = 61 \), then \( 61(n - 1) = 10065 \Rightarrow n - 1 = 165 \Rightarrow n = 166 \)

Thus, there are 3 possible triples \((p, q, n)\) for \( p = 5 \).

### Case 2: \( q = 5 \)
Substitute \( q = 5 \) into the equation:
\[
n \cdot p \cdot 5 = 5(2013p + 5)
\]
Simplify:
\[
5np = 5(2013p + 5)
\]
Divide both sides by 5:
\[
np = 2013p + 5
\]
Rearrange to solve for \( n \):
\[
np - 2013p = 5
\]
Factor out \( p \):
\[
p(n - 2013) = 5
\]
Since \( p \) is a prime number, the only possible value for \( p \) is 5. For this value, we can find \( n \):
- If \( p = 5 \), then \( 5(n - 2013) = 5 \Rightarrow n - 2013 = 1 \Rightarrow n = 2014 \)

Thus, there is 1 possible triple \((p, q, n)\) for \( q = 5 \).

### Case 3: \( p = q = 5 \)
Substitute \( p = 5 \) and \( q = 5 \) into the equation:
\[
n \cdot 5 \cdot 5 = 5(2013 \cdot 5 + 5)
\]
Simplify:
\[
25n = 5(10065 + 5)
\]
Divide both sides by 5:
\[
5n = 10070
\]
Divide both sides by 5:
\[
n = 2014
\]

Thus, there is 1 possible triple \((p, q, n)\) for \( p = q = 5 \).

### Conclusion
Summing up all the possible triples from each case, we have:
- 3 triples from Case 1
- 1 triple from Case 2
- 1 triple from Case 3

Therefore, the total number of triples \((p, q, n)\) is:
\[
3 + 1 + 1 = 5
\]

The final answer is \(\boxed{5}\)