Pigeonhole principle proof sequence of 1's divisible by a number

Question

this problem is a very famous pigeonhole principle problem.

Going from this, how can I prove that 1, 11, 111, ..., 111...111 (2023 ones) has at least one number divisible by 2023?

I began by thinking about the remainders. If I divide all of the numbers by 2023, there are 2023 possible remainders (0, 1, ..., 2022). Since there are 2023 possible remainders and 2023 different numbers, there must be at least one "pigeon" (number) in each "pigeonhole" (remainders). One of which is 0 which represents a number that can be divisible by 2023.

It seems like my reasoning here is flawed. For example, if I take the case of n = 4 and look at the sequence 1, 11, 111, 111, this doesn't hold.

In the original problem, linked at the top, it makes sense that there must be at least one pigeonhole with two pigeons. However, in this case, there are n, not n+1, total numbers. How can I approach this?

Answer 1

The logic isn't that there is a number with remainder $0$. It's that if there is NOT then there are two with same remainder and we can take it from there.

SO..... Either $2023$ divides one of them or it doesn't. If it does we can move along. Nothing to see here folks.

If it doesn't.... then we have $2023$ numbers but only $2022$ non-zero remainders so we have two with the same remainder.

So we have, say $\underbrace{111......1}_n$ and $\underbrace{111......1}_m$ have the same remainder. We can assume $m > n$ and so $\underbrace{111......1}_m -\underbrace{111......1}_n = \underbrace{111......1}_{m-n}\underbrace{0000....0}_n$ is divisble by $2023$.

But $2023$ and $10^n$ are relatively prime. So $2023$ divides $\underbrace{1111.....1}_{m-n}$.

(That's actually a contradict. Either there is or there isn't... and if there isn't there is. So there is.)

.....

If we did that for $4$ we'd conclude that either $4$ divides $1,11,111,$ or $1111$ or that there are at least two with the same remainder and therefore $4$ will divide the difference. Thus we have successfully argued that $4$ must divide one of the following: $1, 11, 111, 1111, 10, 110, 1110, 100, 1100, 1000$. And indeed $4$ divides quite a few (three) of those.

Thing is that $4$ and $10$ are not relatively prime. So knowing that $4|100$ or $4|1100$ or $4|1000$ does not allow us to conclude $4|1$ or $11$.