Let $n$ be an integer which is divisible by neither $2$ nor $5$. Show that $n$ must be divisible by a number whose digits are all $1$.
Consider the following $(n+1)$ integers. $$1, 11, \cdots, \underbrace{11\cdots11}_{n+1}$$
If any of them is a multiple of $n$, then the claim holds. Otherwise, at least two of them must be congruent to each other modulo $n$. Let these two numbers be $\underbrace{11\cdots11}_{m}$ and $\underbrace{11\cdots11}_{k}$ where integers $m > k$. Then we have $$n\mid \underbrace{11\cdots11}_{m} - \underbrace{11\cdots11}_{k} = \underbrace{11\cdots11}_{m-k}\underbrace{00\cdots 00}_{k}$$
We know that $n$ is a multiple of neither $2$ nor $5$. Therefore $$n\mid \underbrace{11\cdots11}_{m-k}\underbrace{00\cdots 00}_{k} \implies n\mid \underbrace{11\cdots11}_{m-k}$$