# Pigeonhole Principle problem

## Homework Statement

Prove that there exists an integer divisible by 1989 Such that it's last four digits are 1990.

## Homework Equations

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.

tiny-tim
Homework Helper
Hi philbein!
Then I need to figure out why it is important that one of these problems have a remainder of one.

Should I add 1989 to the remainder or to the orignal number that I had divided by 1989?

tiny-tim
Homework Helper
erm

if you're not sure, try both!

I'm sorry. I'm still not seeing what this does. Either way, I am still left with a remainder of 1.

matt grime
Homework Helper
You've not quite got it right.

The point is that you have numbers ending in 4 noughts, and when you add 1990 you have a number ending in 1990. You've just got to show...

Ok. I'm still confused.

Dick