A number is called [i]purple[/i] if it can be expressed in the form $\frac{1}{2^a 5^b}$ for positive integers $a > b$. The sum of all purple numbers can be expressed as $\frac{a}{b}$ for relatively prime positive integers $a, b$. Compute $100a + b$.

[i]Proposed by Eugene Chen[/i]
1. **Fix \( b \) and sum over \( a \):**
   We start by fixing \( b \) and summing over all possible values of \( a \) such that \( a > b \). The sum can be written as:
   \[
   \sum_{a=b+1}^\infty \frac{1}{2^a 5^b}
   \]
   Notice that this is a geometric series with the first term \(\frac{1}{2^{b+1} 5^b}\) and common ratio \(\frac{1}{2}\). The sum of an infinite geometric series \(\sum_{n=0}^\infty ar^n\) is given by \(\frac{a}{1-r}\). Here, \(a = \frac{1}{2^{b+1} 5^b}\) and \(r = \frac{1}{2}\). Therefore, we have:
   \[
   \sum_{a=b+1}^\infty \frac{1}{2^a 5^b} = \frac{\frac{1}{2^{b+1} 5^b}}{1 - \frac{1}{2}} = \frac{\frac{1}{2^{b+1} 5^b}}{\frac{1}{2}} = \frac{1}{2^b 5^b} \cdot \frac{1}{2} = \frac{1}{10^b}
   \]

2. **Sum over all \( b \):**
   Now, we sum over all possible values of \( b \) starting from 1:
   \[
   \sum_{b=1}^\infty \frac{1}{10^b}
   \]
   This is another geometric series with the first term \(\frac{1}{10}\) and common ratio \(\frac{1}{10}\). The sum of this series is:
   \[
   \sum_{b=1}^\infty \frac{1}{10^b} = \frac{\frac{1}{10}}{1 - \frac{1}{10}} = \frac{\frac{1}{10}}{\frac{9}{10}} = \frac{1}{9}
   \]

3. **Express the sum in the form \(\frac{a}{b}\):**
   The sum of all purple numbers is \(\frac{1}{9}\). Here, \(a = 1\) and \(b = 9\), and they are relatively prime.

4. **Compute \(100a + b\):**
   \[
   100a + b = 100 \cdot 1 + 9 = 109
   \]

The final answer is \(\boxed{109}\).