2017 Divides String of 1s

Observe that $O_n = \frac{ 10^n - 1 } { 9 }$ . Since $\gcd (9, 2017) = 1$ , hence we are looking for the smallest $n$ such that $10^n \equiv 1 \pmod{2017}$ . Denote this number by $N$ . This is also known as the order of 10 mod 2017.

From Fermat's little theorem , since 2017 is a prime, hence we know that $10 ^ { 2016 } \equiv 1 \pmod {2017}$ . This establishes that $N \leq 2016$ (IE 2016 can be achieved). We will show that $N = 2016$ by proving that no smaller number will work (IE 2016 is a lower bound).

Claim: $N \mid 2016$ .

Proof: Let $2016 = Nq + r$ , where $0 \leq r < N$ .
Then $1 \equiv 10^{2016} \equiv 10^{Nq} \times 10^r \equiv 10 ^ r \pmod{2017}$ . Since $N$ is the smallest positive integer such that $10 ^N \equiv 1 \pmod{2017}$ , this shows that $r$ is not a positive integer so $r = 0$ . Hence $2016 = Nq$ as desired. $_\square$

Claim: $2016 = 2^5 \times 3^2 \times 7$ . If for $M = \frac{2016}{2}, \frac{2016}{3}, \frac{2016}{7}$ , we have $10 ^ M \not \equiv 1 \pmod{2017}$ , then $N = 2016$ .

Proof: Since $N \mid 2016$ , suppose that $N \neq 2016$ . Then $N$ must divide (at least) one of the values of $M$ . For that chosen value, let $M = Nq$ . Then, $1 \equiv 10^{Nq} \equiv 10^m \not \equiv 1 \pmod{2017}$ which is a contradiction. $_\square$

Claim: For $M = \frac{2016}{2}, \frac{2016}{3}, \frac{2016}{7}$ , then $10 ^ M \not \equiv 1 \pmod{2017}$ .

Proof: We have to (tediously) check the values. We obtain:
$10 ^{1008} \equiv 2016 \pmod{2017},$
$10 ^{672} \equiv 294 \pmod{2017},$
$10 ^{288} \equiv 79 \pmod{2017}$ . $_\square$

Note: The above shows that 10 is a primitive root of 2017. Unfortunately, there isn't an easy general way to show that a number is a primitive root, other than checking in the manner that we did.

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(Calvin's rant about minimum)

Remember that when we say $N$ is a minimum, it must satisfy both of the following conditions:
1. It is a lower bound.
2. It can be achieved.

In this case, condition 2 is easy to check, and condition 1 requires work.

$O_n= \overline{111\cdots{ } \cdots{ } \cdots{ }11} \\ =\dfrac{10^n-1}{9}$

Using Euler's Totient Function $\phi (2017)=2016$ as $2017$ is a prime number.

So, $10^{2016} \equiv 1 \pmod{2017}$ as $gcd(10,2017)=1$

Again $10^{2016} \equiv 1 \pmod{9}$

It can be written as $10^n-1$ is both divisible by $2017$ and $9$

Therefore , $10^{2016} -1=9 \times 2017 l$ where $l \in Z^+$

$\dfrac{10^{2016} -1}{9}=2017 l \implies 2017 | \dfrac{10^{2016} -1}{9}$

So, the smallest $n=2016$

Having a look at other solutions and comments, I got to know about a counter example of 11. Indeed, 11 divides (10^2 - 1) / 9 but it is as much fault of the power of 10 as it is of 11 itself. You see, 11 = 10 + 1 and that opens a whole new chain of divisibility taking advantage of the relationship between the terms (a^n - 1) and (a + 1).

It is safe to assume that 11 is an exception rather than the rule. Further examination required.