For some positive integers $m>n$, the quantities $a=\text{lcm}(m,n)$ and $b=\gcd(m,n)$ satisfy $a=30b$.  If $m-n$ divides $a$, then what is the value of $\frac{m+n}{b}$?

[i]Proposed by Andrew Wu[/i]
1. Given the quantities \( a = \text{lcm}(m, n) \) and \( b = \gcd(m, n) \) satisfy \( a = 30b \). We need to find the value of \( \frac{m+n}{b} \) under the condition that \( m-n \) divides \( a \).

2. Recall the relationship between the least common multiple and greatest common divisor:
   \[
   \text{lcm}(m, n) \cdot \gcd(m, n) = m \cdot n
   \]
   Substituting \( a \) and \( b \) into this equation, we get:
   \[
   a \cdot b = \text{lcm}(m, n) \cdot \gcd(m, n) = m \cdot n
   \]
   Given \( a = 30b \), we substitute to find:
   \[
   30b \cdot b = m \cdot n \implies 30b^2 = m \cdot n
   \]

3. We need to find integers \( m \) and \( n \) such that \( m > n \), \( m \cdot n = 30b^2 \), and \( m - n \) divides \( 30b \).

4. Consider the factor pairs of \( 30b^2 \):
   \[
   (m, n) = (6b, 5b)
   \]
   Check if \( m - n \) divides \( 30b \):
   \[
   m - n = 6b - 5b = b
   \]
   Since \( b \) divides \( 30b \), this pair satisfies the condition.

5. Verify other possible pairs:
   \[
   (m, n) = (10b, 3b) \implies m - n = 7b \quad (\text{does not divide } 30b)
   \]
   \[
   (m, n) = (15b, 2b) \implies m - n = 13b \quad (\text{does not divide } 30b)
   \]
   \[
   (m, n) = (30b, b) \implies m - n = 29b \quad (\text{does not divide } 30b)
   \]

6. The only valid pair is \( (m, n) = (6b, 5b) \). Calculate \( \frac{m+n}{b} \):
   \[
   \frac{m+n}{b} = \frac{6b + 5b}{b} = \frac{11b}{b} = 11
   \]

The final answer is \(\boxed{11}\)