For the problem seehttps://learn1.open.ac.uk/mod/oublog/viewpost.php?post=242384
This was new to me. I found it in Mathematical Puzzles: A Connoisseur’s Collection, by Peter Winkler.
Think of the 2023 numbers 1, 11, 111, 1111, … up to the number with 2023 digits, all 1’s. Imagine dividing each by 2022 and taking the remainder. There are only 2022 distinct remainders possible 0, 1, 2, 3, … 2021 but we have produced 2023 remainders, and thus there must be at least two of the numbers 1, 11, 111, 1111, … with the same remainder.
If we now take the smaller of the two from the larger, the number we get must be divisible by 2022, and it will consist only of 0’s and 1’s .as required.
It’s late, hope I have the details right, but you should be able to see that the argument is correct and 2022 could be replaced by any number we please.