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Let's call the "complexity" of a given number the smallest length of a numerical sequence (if such exists) that starts with zero and ends with this number, where each subsequent member of the sequence is either half of the previous one or, when added to the previous one, equals 1. Among all numbers of the form $m / 2^{50}$, where $m=1,3,5, \ldots, 2^{50}-1$, find the number with the greatest "complexity".
The answer: the maximum "complexity" is achieved by the number $\left(\left(2^{51}+1\right) / 3\right) / 2^{50}$.

We will prove that among numbers of the form $m / 2^k$, where $m=1,3,5, \ldots, 2^k-1$, the maximum "complexity" (equal to $2k$) is achieved by the number for which $m=m_0(k)=\left(2^{k+1}+(-1)^k\right) / 3$. We will prove this statement by induction on $k$.

Base case. $k=1$. Clearly, the "complexity" of the number $1 / 2$ is two.

Inductive step. Assume that our statement is proven for $k$. We will prove it for $k+1$. First, note that the complexity of any number of the form $m / 2^{k+1}$ does not exceed $2(k+1)$. Indeed, if $m < 2^{k+1}$, then this number can be obtained as follows: first, obtain the number $\left(2^{k+1}-m\right) / 2^k$, then divide it by two, and then subtract it from one (a total of no more than $2k+1+1=2(k+1)$ operations). Now we will prove that the "complexity" of the number $m_0(k+1) / 2^{k+1}$ is $2k+2$. Indeed, since $m_0(k+1) / 2^{k+1} > 1 / 2$, the last operation in obtaining this number is subtraction from one, and the second-to-last operation is division by two. But this sequence of operations results in the number $m_0(k) / 2^k$. Therefore, the "complexity" of the number $m_0(k+1) / 2^{k+1}$ is two more than the "complexity" of the number $m_0(k) / 2^k$, i.e., it is $2k+2$, Q.E.D.