Serious question: Prove that for any positive integer n there exists a positive multiple of n that contain only the digits 7 and 0. (in decimal system, of course.)
I know I'm supposed to use some sort of the pigeonhole principle, but I have no idea what the pigeons and pigeonholes are. Any hints?

Question

Serious question: Prove that for any positive integer n there exists a positive multiple of n that contain only the digits 7 and 0. (in decimal system, of course.)
I know I'm supposed to use some sort of the pigeonhole principle, but I have no idea what the pigeons and pigeonholes are. Any hints?

experimentx · Answer

1*7 = 7
2*35 = 70
3*259=777
4*175=700
5*14  = 70
6*1295 = 7770
7*10 = 70
..
..
looks like we really can construct it.

thomas5267 · Answer

This is the pigeonhole principle, but I have no idea how to use this to prove the question. http://en.wikipedia.org/wiki/Pigeonhole_principle

experimentx · Answer

I hate to do it ... but there is answer in google ... take a peek if you like!!! :(

anonymous · Answer

Looked at the solution on google, and it makes sense, but I would never have figured that out myself.

experimentx · Answer

me ... not in next 100 years!!!

blockcolder · Answer

My solution was close, but I didn't realize I could just subtract the 2 numbers in the same pigeonhole. I thought I had to prove that one of pigeons should lie in the pigeonhole labeled 0.
Anyway, thanks for telling me about the internet solution.
This is what I get for enrolling in a problem-solving course.
Closing in 3, 2, 1...