Suppose $f(n)$ is the minimum time it takes to get $n$ adult worms starting with one adult worm. Observe that $f(1) = 0$. Suppose we cut the initial worm into smaller worms of size $x$ and $1-x$, where $x \in (0, 1)$. If the $x$ length worm, after maturing grows into exactly $k$ worms; then the other one must grow into at least $n-k$ worms, implying that $f(n) \geq \min_{1 \leq k < n, x \in (0, 1)} \max(1-x+f(k), x+f(n-k))$. Fixing some $k$, the optimum value of $x$ is obviously when the two quantities involved are equal; hence $f(n) \geq \min_{1 \leq k \leq n} \frac{1+f(k)+f(n-k)}{2}$. It is easy to see that this is achievable as well, hence the inequality is in fact an equality. From here, a simple induction shows that $f(n) = 1-\frac{1}{2^{n-1}}$ for all $n \geq 1$. Hence the minimum time required to grow $n$ adult worms starting with only one is equal to $1-\frac{1}{2^{n-1}}$.