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
Best Answer
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 $n$.