Let $A, B$, and $F$ be positive integers, and assume $A < B < 2A$. A flea is at the number $0$ on the number line. The flea can move by jumping to the right by $A$ or by $B$. Before the flea starts jumping, Lavaman chooses finitely many intervals $\{m+1, m+2, \ldots, m+A\}$ consisting of $A$ consecutive positive integers, and places lava at all of the integers in the intervals. The intervals must be chosen so that:

([i]i[/i]) any two distinct intervals are disjoint and not adjacent;
([i]ii[/i]) there are at least $F$ positive integers with no lava between any two intervals; and
([i]iii[/i]) no lava is placed at any integer less than $F$.

Prove that the smallest $F$ for which the flea can jump over all the intervals and avoid all the lava, regardless of what Lavaman does, is $F = (n-1)A + B$, where $n$ is the positive integer such that $\frac{A}{n+1} \le B-A < \frac{A}{n}$.
1. **Define Variables and Initial Setup:**
   Let \( A, B, \) and \( F \) be positive integers such that \( A < B < 2A \). Define \( C = B - A \). The flea can jump to the right by \( A \) or \( B \). Lavaman places lava in intervals of \( A \) consecutive integers, ensuring the intervals are disjoint and not adjacent, with at least \( F \) positive integers between any two intervals, and no lava at any integer less than \( F \).

2. **Part 1: Prove \( F = nA + C \) is sufficient:**
   - **Safe and Lava Sections:**
     Consider a safe section of length \( nA + C + R \) where \( 0 \le R < A \). Redefine the number line so that \( 1 \) is the beginning of this safe section. We need to show that from any of the first \( C \) positions, the flea can move to the first \( C \) positions of the next safe section.

   - **Jump Sequences:**
     Let \( R = kC + r \) where \( k \) and \( r \) are nonnegative integers with \( k \) maximal. Notice that \( n + 1 > k \) since \( \frac{R}{n+1} < \frac{A}{n+1} \le C \). Consider the following sequences of moves:
     \[
     [r + 1, C] \stackrel{+(n-k)A + kB}{\to} nA + kC + [r + 1, C] = nA + R + [1, C-r]
     \]
     and
     \[
     [1, r] \stackrel{+(n-k-1)A + (k+1)B}{\to} nA + kC + [1, r] = nA + R + [1 + C-r, C] \quad \text{if } n \ge k + 1.
     \]
     \[
     [1, r] \stackrel{+(n+1)A}{\to} nA + A + [1, r] = nA + R + [A-R+1, A-nC] \quad \text{if } n = k.
     \]
     These final positions lie in the interval \( nA + R + [1, C] \), fitting into the safe section. Another leap of length \( B = A + C \) gives the desired result.

3. **Part 2: Prove \( F = nA + C \) is necessary:**
   - **Safe and Lava Sections:**
     Divide the number line into safe and lava sections. If Lavaman makes every safe section of length \( nA + C - 1 \), the flea cannot traverse \( C \) lava intervals. If the flea is on some position in \( [s, C] \) for some \( s \le C \), it must pass through \( [s+1, C] \) or move no more.

   - **Jump Analysis:**
     The leap over the lava must be of length \( B = A + C \), so the flea must be in \( [nA, nA + C - 1] \) to make this jump. If the flea moves from \( [s, C] \) to \( nA + [0, C-1] \) in \( n-1 \) jumps, we need:
     \[
     (n-1)(A+C) \ge nA - C \implies A \le nC \implies B - A \ge \frac{A}{n},
     \]
     which is a contradiction. Hence, at least \( n \) jumps are needed. \( n \) jumps of length \( A \) send \( [s, C] \) to \( nA + [s, C] \), so the flea must do exactly \( n \) jumps of length \( A \). Now \( nA + C \) is lava, so if the flea was at \( C \) initially, it cannot pass the lava. Otherwise, the flea jumps \( B = A + C \) to \( (n+1)A + C - 1 + [s + 1, C] \).

   - **Conclusion:**
     The flea starts at position \( 1 \) relative to its safe section within \( [1, C] \). After \( C \) lava intervals, the flea runs out of room. Hence, \( F = nA + C = (n-1)A + B \) is minimal.

\(\blacksquare\)

The final answer is \( \boxed{ F = (n-1)A + B } \).