You drop a 7 cm long piece of mechanical pencil lead on the ﬂoor. A bully takes the lead and breaks it at a random point into two pieces. A piece of lead is unusable if it is 2 cm or shorter. If the expected value of the number of usable pieces afterwards is $\frac{m}n$ for relatively prime positive integers $m$ and $n$, compute $100m + n$.

[i]Proposed by Aaron Lin[/i]
1. Let's denote the length of the mechanical pencil lead as \( L = 7 \) cm. The lead is broken at a random point \( X \) where \( 0 < X < 7 \).

2. We need to determine the probability that each piece of the broken lead is usable. A piece is considered usable if its length is greater than 2 cm.

3. The left piece has length \( X \) and the right piece has length \( 7 - X \).

4. For the left piece to be usable, we need \( X > 2 \). The probability of this event is:
   \[
   P(X > 2) = \frac{7 - 2}{7} = \frac{5}{7}
   \]

5. For the right piece to be usable, we need \( 7 - X > 2 \), which simplifies to \( X < 5 \). The probability of this event is:
   \[
   P(X < 5) = \frac{5 - 0}{7} = \frac{5}{7}
   \]

6. By the linearity of expectation, the expected number of usable pieces is the sum of the expected usability of each piece. The expected usability of the left piece is \( \frac{5}{7} \) and the expected usability of the right piece is \( \frac{5}{7} \).

7. Therefore, the expected number of usable pieces is:
   \[
   \frac{5}{7} + \frac{5}{7} = \frac{10}{7}
   \]

8. The expected value of the number of usable pieces is \( \frac{10}{7} \), where \( m = 10 \) and \( n = 7 \).

9. To find \( 100m + n \):
   \[
   100 \cdot 10 + 7 = 1000 + 7 = 1007
   \]

The final answer is \( \boxed{ 1007 } \).