A worm is called an [i]adult[/i] if its length is one meter. In one operation, it is possible to cut an adult worm into two (possibly unequal) parts, each of which immediately becomes a worm and begins to grow at a speed of one meter per hour and stops growing once it reaches one meter in length. What is the smallest amount of time in which it is possible to get $n{}$ adult worms starting with one adult worm? Note that it is possible to cut several adult worms at the same time.
1. Define \( f(n) \) as the minimum time required to get \( n \) adult worms starting with one adult worm. Clearly, \( f(1) = 0 \) because we start with one adult worm.
2. Consider cutting the initial adult worm into two parts of lengths \( x \) and \( 1-x \), where \( x \in (0, 1) \). Each part will grow to become an adult worm in time proportional to its length.
3. Suppose the worm of length \( x \) grows into \( k \) adult worms. Then, the worm of length \( 1-x \) must grow into \( n-k \) adult worms. The time required for this process is the maximum of the times required for each part to grow into the respective number of worms.
4. Therefore, we have:
   \[
   f(n) \geq \min_{1 \leq k < n, x \in (0, 1)} \max(1-x + f(k), x + f(n-k))
   \]
5. To find the optimal \( x \) for a fixed \( k \), we set the two quantities equal:
   \[
   1 - x + f(k) = x + f(n-k)
   \]
   Solving for \( x \), we get:
   \[
   x = \frac{1 + f(n-k) - f(k)}{2}
   \]
6. Substituting this back, we get:
   \[
   f(n) \geq \min_{1 \leq k < n} \frac{1 + f(k) + f(n-k)}{2}
   \]
7. By induction, we can show that \( f(n) = 1 - \frac{1}{2^{n-1}} \) for all \( n \geq 1 \). 

   **Base Case:** For \( n = 1 \), \( f(1) = 0 \), which matches \( 1 - \frac{1}{2^{1-1}} = 0 \).

   **Inductive Step:** Assume \( f(k) = 1 - \frac{1}{2^{k-1}} \) holds for all \( k < n \). Then:
   \[
   f(n) = \min_{1 \leq k < n} \frac{1 + f(k) + f(n-k)}{2}
   \]
   Substituting the inductive hypothesis:
   \[
   f(n) = \min_{1 \leq k < n} \frac{1 + \left(1 - \frac{1}{2^{k-1}}\right) + \left(1 - \frac{1}{2^{(n-k)-1}}\right)}{2}
   \]
   Simplifying:
   \[
   f(n) = \min_{1 \leq k < n} \frac{2 - \frac{1}{2^{k-1}} - \frac{1}{2^{n-k-1}}}{2}
   \]
   \[
   f(n) = 1 - \frac{1}{2} \left( \frac{1}{2^{k-1}} + \frac{1}{2^{n-k-1}} \right)
   \]
   The minimum value occurs when \( k = 1 \) or \( k = n-1 \), giving:
   \[
   f(n) = 1 - \frac{1}{2^{n-1}}
   \]

Thus, the minimum time required to grow \( n \) adult worms starting with only one is \( 1 - \frac{1}{2^{n-1}} \).

The final answer is \( \boxed{ 1 - \frac{1}{2^{n-1}} } \).