Show that for every integer $n$ there is a multiple of $n$ that has only $0s$ and $1s$ in its decimal expansion. Can anyone please explain this example as I tried a lot to understand it but I can't!
The problem:

Show that for every integer n there is a multiple of n that has only
0s and 1s in its decimal expansion.

The Solution of the book:

Let $n$ be a positive integer. Consider the $n + 1$ integers $1, 11,$
$111, ..., 1111, ...$ (where the last integer in this list is the
integer with $n + 1$ $\ 1s$ in its decimal expansion). Note that there
are $n$ possible remainders when an integer is divided by $n$. Because
there are $n + 1$ integers in this list, by the pigeonhole principle
there must be two with the same remainder when divided by $n$. The
larger of these integers less the smaller one is a multiple of $n$,
which has a decimal expansion consisting entirely of $0s$ and $1s$.

This problem from Discrete Mathematics and its application's for Rosen
 A: Suppose, say, that $n=3$. Consider the four numbers $1$, $11$, $111$, and $1\,111$. What are the remainders of the division of these numbers by $3$? They are $1$, $2$, $0$, and $1$ respectively. The remainder $1$ appears twice (corresponding to the numbers $1$ and $1\,111$). So, $1\,111-1(=1\,110)$ is a multiple of $3$.
The same idea works with every $n$.
A: Here's a way I used to explain this to myself -
Let n be a positive integer. Let's consider n+1 numbers of form 1, 11, 111, 1111, 1111 .... where last integer in this list is an integer with n+1 1s.
By the pigeonhole principle, there must be at least 2 numbers in this list of n+1 numbers that have the same remainder when divided by n.
Let these numbers be a & b.
a = 1...1 = Q.n + r, where Q is some integer and r is some integer < n
b = 1.....1 - Q'.n + r, where Q' is some integer and r is some integer < n
(r is common remainder)
b > a
b - a = 111...11...11 - 11..111 = 111...00..000 = n(Q' - Q) + r - r = n.(Q' - Q)
=> b - a is a multiple of n that has only 0s and 1s in it's decimal.
