<s>[INST] Finding the number of digits in repunit Find the number of digits in the smallest repunit divisible by $19$.
I believe a repunit number, with $N$ digits, is of the form $ \sum_{k=0}^{N-1} 10^k = \frac{10^N -1}{10-1} = \frac{10^N - 1}{9}, $ and is divisible by $n$ if $10^N - 1 \equiv 0 \pmod{9n}.$ The solution should be the smallest number N that satisfies $10^N -1 \equiv 0 \pmod{9 \cdot 19}.$ Can somebody finish this off?
 [/INST] $10\equiv1\pmod9$, so $10^N\equiv 1 \pmod9$ for all $N\in\mathbb N$,
so your question becomes what is the smallest number $N$ satisfying $10^N\equiv1\pmod{19}$.
By Fermat's little theorem, we know $10^{18}\equiv1\pmod{19}$;
we just have to show that $10^6\not\equiv1\pmod{19}$ and $10^{9}\not\equiv1\pmod{19}$.
Method 1
$10^2\equiv5\bmod19$, so $10^3\equiv50\equiv12\bmod19$, so $10^6\equiv144\equiv11$, and $10^9\equiv132\equiv18\bmod19$.
Method 2
$10^6-1=(10^3+1)(10^3-1)=(7\times11\times13)(27\times37)$ is not divisible by $19$.
By Euler's criterion,  $10^9-1\equiv\left(\dfrac{10}{19}\right)=\left(\dfrac{2}{19}\right)\left(\dfrac{5}{19}\right)=(-1)\left(\dfrac45\right)=-1\pmod{19}$.
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