Prove there is an integer divisible by 2003

Question

I don't quite understand why $s_j$ $-$ $s_j$ is divisible by $2003$ means that there is an element in this sequence that is divisible by $2003$. Aren't they trying to prove for one single element? Not for a difference of the sequence's two elements?

Answer 1

Because .... part i)

Just read part ii) first and then part i).

PART II

A statement of contradiction. If $2003$ doesn't divide any $s_k$ then there are only $2002$ possible remainders for $s_k$ divided by $2003$. By Pigeon hole two have the same remainder. Let $s_j$ be the larger and $s_i$ be the smaller. So $s_j - s_i$ has remainder $0$ and $2003$ divides $s_j - s_i$.

PART II

$s_j - s_i = 10^i*s_{j-i}$ and $2003$ is prime and does not divide $10^i$. So if $2003|s_j - s_i$ then $2003|s_{j-i}$.

.....

So proof by contradiction:

If $2003$ doesn't divide any $s_k$ then by PART I, we have that there are $s_j$ and $s_i$ so that $2003|s_j - s_i$ and by PART II we have $2003|s_{j-i}$.

That's a contradiction so $2003$ divides some $s_k$.