Period of repeating decimals

Solution 1:

By period, I assume you mean minimal period, otherwise there are inifinitely many periods.

Write $i=x_iy_i$, where every prime factor of $x_i$ divides $b$, and $\gcd(y_i, b)=1$. Then $f(i)=f(y_i)=ord_{y_i}(b)$, where $ord_{y_i}(b)$ is the multiplicative order of $b \pmod{y_i}$.

This is because:

$x_i$ does not affect the minimal period $f(i)$.

If $y_i \mid (b^n-1)$ for some $n$, then let $b^n-1=ky_i$, then $$\frac{1}{y_i}=\frac{k}{b^n-1}=\frac{k}{b^n(1-b^{-n})}=\frac{k}{b^{n}}+\frac{k}{b^{2n}}+ \ldots$$

So $\frac{1}{y_i}$ has $n$ as a period. Since the smallest such $n$ is given by $ord_{y_i}(b)$, $f(y_i) \leq ord_{y_i}(b)$.

On the other hand, since $f(i)$ is a period, let $l$ be the number representing the repeating string, so we can write $$\frac{1}{y_i}=\frac{l}{b^{f(i)}}+\frac{l}{b^{2f(i)}}+ \ldots=\frac{l}{b^{f(i)}(1-b^{-f(i)})}$$

Thus $b^{f(i)}-1=ly_i$, so $f(y_i) \geq ord_{y_i}(b)$.

Now back to your results.

1 is mostly correct, as we have $f(i)=f(y_i)=ord_{y_i}(b) \leq \varphi(y_i) \leq y_i-1 \leq i-1$ for $y_i>1$, and if $y_i=1$, $f(i)=1 \leq i-1$ for $i>1$. The only exception is $i=1$, as $f(i)=1>1-1$.

2 is partially correct. The $b^n-1$ part is correct, but not the $b^{n-1}$ part.

3 is correct. $y_i=y_{j_1}y_{j_2}$ so $f(i)=ord_{y_i}(b)=lcm(ord_{y_{j_1}}(b),ord_{y_{j_2}}(b))=lcm(f(j_1),f(j_2))$.