Binary Numbers! (Though Not Really)

Relevant wiki: Pigeonhole Principle - Problem Solving

Firstly, if the number $n$ is of the form $2^a \cdot 5^b$ , then $10^{\max (a, b)}$ is a multiple of $n$ .

If $n$ is odd and not divisible by 5, consider the following:

$\underbrace{1, 11, 111, \ldots}_{n+1 \text{ numbers}}$

By Pigeonhole Principle, at least 2 numbers share the same residue in $\pmod{n}$ . Let these numbers have $i$ and $j$ digits, with $i>j$ . We have

$\begin{aligned} \underbrace{111...}_{i} - \underbrace{111...}_{j} &\equiv 0 \pmod{n}\\ \underbrace{111...}_{i-j} \underbrace{000...}_{j} &\equiv 0 \pmod{n}\\ \underbrace{111...}_{i-j} &\equiv 0 \pmod{n} \end{aligned}$

Thus, there does exist a multiple that satisfies.

If $n$ is not in either of the above categories, we do as follows:

Let $a$ and $b$ be non-negative integers such that $2^a \cdot 5^b \mid \mid n$ . Firstly, we find the multiple which satisfies the conditions for $\frac {n}{2^a 5^b}$ , then we multiply this number by $10^{\max(a, b)}$ . This number will also satisfy.

Thus, all numbers have a multiple composed entirely of 1's and 0's.