An interesting problem with binary repetend Proof. Let $b$ is an odd number, Then $1/b$ in the binary loop section, the number of $1$ does not exceed the number of $0$. 
for example:
$1/3=0.010101……$
$1/5=0.001100110011……$
$1/7=0.001001001……$
$1/9=0.000111000111……$
We know that satisfy the congruence expression $2^d≡1\ (mod \ b)$ of the Least positive integer $d$ , it's $d$ that $1/b$ of the repetend longth.
In particular, when $b-d=1$, the number of $0$ is equal to the number of $1$.
 A: Unfortunately, this does not seem to be true.
In binary, we have that
$$
  \frac{1}{187} = 0.\overline{0000001010111100111010110111011100011010},
$$
and the repeating part contains $19$ zeroes, and $21$ ones.
A: As Dylan's answer shows, this is not true.  Let me give a bit of an explanation for why it's not surprising that counterexamples are rare.  Note that if the repetend has length $d$, then it is just the binary expansion of $\frac{2^d-1}{b}$, since $\frac{1}{b}=\frac{(2^d-1)/b}{2^d-1}$.  Roughly speaking, you might expect that $\frac{2^d-1}{b}$ is an "average" number of its size.  This means that typically, it will contain about $\frac{1}{2}\log_2\frac{2^d-1}{b}\approx\frac{d}{2}-\frac{\log_2 b}{2}$ digits which are $1$.  Because of the $-\frac{\log_2 b}{2}$ term, this is less than $\frac{d}{2}$, so the repetend needs to have an unusually large number of $1$s in order to have more $1$s than $0$s.  To put it another way, the digits of the repetend must start with a bunch of $0$s (namely, $\log_2 b$ of them), and so the rest of repetend must have an unusual excess of $1$s to make up this deficit.
Moreover, there is a class of cases (which is fairly common at least for small $b$) in which exactly half of the repetend is $1$s.  Namely, if there exists $n$ such that $2^n=-1$ mod $b$, then $d=2n$ and the repetend consists of a binary string of length $n$ followed by the same binary string but with all of its digits flipped.  Indeed, the $k$th digit of $1/b$ is determined by whether the remainder of $2^{k-1}$ mod $b$ is greater than or less than $b/2$, and so increasing $k$ by $n$ will always flip the digit.
