Prove that there exists an integer divisible by 1989 Such that it's last four digits are 1990.
Pigeonhole Principle where if we have k+1 items and k holes, two items go in one hole.
The Attempt at a Solution
Ok. I called my professor for help on this and I was given the hint that I should first take the integers from 1 to 1989, and multiply each one of them by 10^4. Then, when I divide each number, each has a different remainder, which means one of them has a remainder of one. I first need to figure out why this whole thing is true ( meaning why they each have different remainders). Then I need to figure out why it is important that one of these problems have a remainder of one. I have no clue on what to do here, so any suggestions would be great. Thanks for your time.