Primes dividing $11, 111, 1111, ...$ [duplicate]

Hint: Use Fermat's little theorem to conclude that $10^{n(p-1)} - 1$ is divisible by $p$ for every prime $p$ other than $2, 5$.

Define the infinite sequence $S =s_2, s_3, s_4 \dots $ as $11, 111, 1111, \dots$.

Consider all $s_i$ modulo a prime $p$ other than 2 or 5. By the pigeonhole principle, there must be a residue $r$ such that infinite $s_i$ are congruent to it. Let $s_a$ be the smallest such element of $S$.

Let $s_b$ be one of the infinite set of elements of $S$ with residue $r$ that are greater than $s_a$. Their difference, $s_b - s_a$ is divisible by $p$. It has the form $11\dots1100\dots000 = s_{b-a} \cdot 10^{a}$. Since $p$ is neither 2 nor 5, $p$ must divide $s_{b-a}$. Since there is an infinite set of $s_b > s_a$ with residue $r$, there must also be an infinite set of $s_{b-a}$ divisible by $p$.