0

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?

72inches
  • 79
  • 2
    Hint: these numbers are $\frac19(10^n-1)$ for $1\le n\le2023$. What congruence needs to hold for $10^n$ modulo $2023$ to make such a number a multiple of $2023$? – Greg Martin Nov 13 '22 at 21:29
  • Another hint: Consider the remainders when $1, 11, 111, 1111, \cdots $ are divided by 2023. What does the pigeinhole principle tells us? – Li Kwok Keung Nov 13 '22 at 23:18
  • You also need to check your understanding of the PHP. It doesn't guarantee that each hole contains a pigeon, it guarantees that at least one hole has two or more pigeons in it. To apply the principle to this problem, you would need to construct a set of items and divide them into classes such that two items belonging to the same class forces 1...1 to be divisible by 2023. – ConMan Nov 13 '22 at 23:21
  • 1
    Dupe $ $ Special case of answers in Prove that every number ending in a $3$ has a multiple which consists only of ones., e.g. see this answer *as well as many others dupe targets). – Bill Dubuque Nov 14 '22 at 08:47
  • 3
    I don't think it's a dupe as the OPs question was not understanding a specific line of a solution. Reading a thousand answers won't help the OP if they don't get the line. On the other hand explaining a single line doesn't warrant a full answer among dozens of dupes. ... Maybe if there were some classification between responsed and "big" answers vs. "small" answers or something. – fleablood Nov 14 '22 at 16:18
  • @LiKwokKeung I understand that if there are 2023 + 1 of those numbers in the sequence of 1s, the PHP states that there must be at least 1 hole with 2 pigeons (numbers) in it. However, in my case, I am only considering 2023 numbers and trying to prove that at least one of them are divisible by 2023. I am lost in trying to apply the PHP to this. – khkh Nov 14 '22 at 22:55
  • 1
    @khkh fleablood had explained it. Upon division by 2023, there are 2023 possible remainders, namely $0, 1, 2, \cdots, 2022$. You have made 2023 divisions, and thus got 2023 remainders. If one of your remainder is $0$, you are done. If you do not get $0$ as one of the remainders, then your 2023 remainders can only have 2022 choices, i.e. from $1$ up to $2022$. PHP tells us that at least $2$ of your (2023) remainders must be the same. Then fleablood's argument applies. – Li Kwok Keung Nov 14 '22 at 23:39

1 Answers1

3

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 $11$.

fleablood
  • 124,253