Is it true, that every prime (except 2) can be found as a divisor of enough long series of 1-s? I suspect, that the prime divisors of the series of 1, 11, 111, 1111, ... 1...1 will contain every primes. But this is an intuitive hypothesis only. Somebody knows, if it is so? How could it be proven/disproven?
 A: We can prove more very simply: if $\rm\:m\:$ is coprime to $10\,$ then any number with $\rm\: m\:$ digits all $\ne 0$  has a contiguous digit subsequence that forms a number divisible by $\rm\:m.\:$ Suppose the digits are $\rm\:d_{m}\ldots d_1.\:$ By $\rm\,d_i\ne 0\:$ the $\rm\:m\!+\!1\:$ numbers $\rm\:0,\,d_1,\, d_2 d_1,\, d_3 d_2 d_1,\, \ldots,d_m\!\ldots d_1$ are distinct. By Pigeonhole two  are congruent $\rm\:mod\ m,\:$  so $\rm\:m\:$ divides their difference $\rm = 10^k\:$ times the number $\rm\,n\ne 0\,$ formed by the extra digits of the longest, so $\rm\:m\,|\,10^kn\:$ $\Rightarrow$ $\rm\:m\:|\:n,\:$ by $\rm\:m\:$ is coprime to $10.\:$
Let's do a simple example. Let $\rm\:m=9,\:$ and let the number be $\rm\:98765\color{green}{432}1.\:$ Modulo $9$ we have $\rm\:1 \equiv\color{#C00} 1,\ 21\equiv 3,\ 321\equiv 6,\ 4321\equiv\color{#C00} 1,\:$  so $\rm\:9\:|\:4321\!-\!1 = 432\cdot 10,\:$ so $\,9\,|\,\color{green}{432}.$
In your case, the divisor $\rm\:m=p\:$ is a prime coprime to $10$, so the number $\,111\ldots 111$ $\rm\,(m$ digits) does the trick, i.e. some subsequence $11\ldots 11$ is divisible by $\rm\:m.$
The result extends to any number having $\rm\:m\:$ nonzero digits: simply take the subsequences beginning with nonzero leading digit. This implies that the $\rm\:m+1\:$ numbers are increasing (so distinct), and the number formed by the extra digits is nonzero, since its leading digit is nonzero.
A: Tobias Kildetoft's excellent answer (posted as comment) was this:


*

*Consider any prime $p$ apart from 2 or 5.

*There are infinite many numbers like 1...1. So there must be two between them having the same remainder $\mod p$. Call them for example $a$ and $b$.

*$b-a$ will have to look like 1...10...0, and it will be divisible by $p$.


QED.
A: Consider the sequence $10^k$ modulo $p$ (except for $p=2,3,5$, which we will handle separately). Since there are only $p$ equivalence classes mod $p$, for any group of $p+1$ $k$'s, we must have $k_1\gt k_2$ so that $10^{k_1}\equiv10^{k_2}\pmod{p}$. Since $(p,10)=1$, this implies
$$
10^{k_1-k_2}-1\equiv0\pmod{p}
$$
and since $(p,3)=1$, we have
$$
\frac{10^{k_1-k_2}-1}9\equiv0\pmod{p}
$$
This is a sequence of $k_1-k_2$ ones in base-ten.
$$
111\equiv0\pmod3
$$
This is not true for $p=2$ and $p=5$ since no multiple of $2$ or $5$ can be equal to $1$ mod $10$.
