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

Home > Just Math
Final Minus Two (Posted on 2015-01-24) Difficulty: 3 of 5
Start with the set S = {0, 2014}. Then, repeatedly, expand S as follows.

Place into S any integer that is a root of a polynomial the coefficients of which are in S.

Prove that the negative number −2 eventually appears in S.

No Solution Yet Submitted by Danish Ahmed Khan    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution What am I missing this time (spoiler?) | Comment 1 of 2
x = -2 is a root of the polynomial x + 2 = 0.

So, it seems like x = -2 gets added during the first expansion of S.

I must be missing something, if this problem is difficulty 3.

----------------------------------------------------------

Oh, now I see.  S starts with only two elements, 0 and 2014.  Back to the drawing board.


Edited on January 24, 2015, 1:57 pm
  Posted by Steve Herman on 2015-01-24 13:52:27

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 (4)
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