The length of the smallest non-repeating block in decimal representation of $\frac{1}{p}$ , where $p \in \mathbb{P} \setminus \{ 2,3,5\} $ is always even. Proof : Let $p$ be a prime and $ p\not=2,3,5$ let $$ \frac{1}{p} = 0.\overline{ a_{k-1}a_{k-2}\dots a_1a_0 } $$where $a_i \in \{0,1,2,\dots,9\} \ \forall \ 0 \leq i \leq n $ So the length of the smallest non-repeating block in decimal representation of $\frac{1}{p}$ is $k$. $$ \therefore \frac{10^k - 1}{p} = a_{k-1}a_{k-2}\dots a_1a_0 \in \mathbb{Z} $$$\Rightarrow k$ is the minimum positive integer such that $ p | 10^k - 1 $ Thus $ k = ord_p(10) | p-1 \Rightarrow k $ is even ( as $k \not= 1, \ \because p \not= 3$ )

Let $N = ( b_t b_{t-1} \dots b_1b_0 )_{10}$ then for all positive integer $ v > t $ we have $ S(Nx) = 9v $ where $ x = 10^v - 1 $ Proof : $$ Nx = ( b_t b_{t-1} \dots b_1 b_0 )_{10} \left( \underbrace{ 99\dots 99 }_{ v \mbox{ times } 9 } \right) $$Assume that $b_0 \not= 0$ because if $b_0 = 0$ we can simply look at $\frac{Nx}{10} $ Now consider the number $$ \alpha = \left( b_t b_{t-1} \dots b_1 ( b_0 - 1) \underbrace{99\dots99}_{ v-t-1 \mbox{ times } 9 }(9-b_t)(9-b_{t-1}) \dots (9-b_1)(9-b_0 +1 ) \right)_{10} $$We will prove that $ \alpha = Nx $ $$ \alpha = \left( b_t \dots b_1( b_0-1) \right)_{10} \cdot 10^v + \left( \underbrace{99\dots99}_{v-t-1 \mbox{ times } 9 } \right) \cdot 10^{t+1} + \left( (9-b_t)(9-b_{t-1})\dots(9-b_1)(9-b_0+1) \right)_{10} $$$$ \Rightarrow \alpha = N \cdot 10^v - 10^v + \left( 10^{v-t-1} -1 \right) \cdot 10^{t+1} + 9 \left( \sum_{i=0}^{t} 10^i \right) - N + 1 $$$$ \Rightarrow \alpha = N \cdot 10^v - 10^v + (10^v - 10^{t+1}) + (10^{t+1} -1) - N+1 = N( 10^v-1) = Nx $$$$ \therefore S(Nx) = S(\alpha) = \left(\sum_{i=0}^t b_i \right) -1 + 9( v-t-1) -\left( \sum_{i=0}^t ( 9 - b_i ) \right) + 1= 9v $$