All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Numbers > Sequences
Many Multiples (Posted on 2012-04-30) Difficulty: 2 of 5
Prove that infinitely many numbers in the sequence {2012, 20122012, 201220122012, 2012201220122012,....} are multiples of 101.

No Solution Yet Submitted by K Sengupta    
Rating: 4.0000 (2 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Egg on my face | Comment 7 of 8 |
Boy, was I wrong.  Wrong about my proof and wrong about the generalization.

Consider the sequence 0, 2, 22, 222, 2222, etc.

It has only one term divisible by 5, not an infinite number.

While the remainders mod 5 do eventually cycle (in this case, with period 1), that does not mean that the first term will be part of that cycle.

And that's where I went wrong. 

  Posted by Steve Herman on 2012-05-01 14:32:42
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (2)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2017 by Animus Pactum Consulting. All rights reserved. Privacy Information