Let $d$ be any fixed natural number. Show that there must exist an integer of the form $11\ldots1100\ldots 00$ (that is a integer whose digits consist of a sequence of $1$'s followed by $0$'s) which is divisible by $d$.
So If $x$ is the sequence integer the I know I want to show that:
$$x \equiv 0 \mod d$$
I know that the remainder are the boxes in the pigeon hole principles. But here I am stuck on how to proceed. any help?
Consider the first $d+1$ terms of this sequence
$$ 1, 11, 111, ...$$ and their remainders mod $d$. By the pigeonhole principle, we know that two of these must have the same remainder modulo $d$. Call them $a_i$ and $a_j$ where
$$ a_i = \underbrace{1111...1}_\textrm{i ones} $$ $$ a_j = \underbrace{1111...1}_\textrm{j ones} $$
and $i < j$. But then,
$$a_j - a_i = \underbrace{1111...1}_\textrm{j-i ones}\underbrace{0000...0}_\textrm{i zeroes}$$
and is divisible by $d$.