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

 Last but not the least (Posted on 2013-12-12)
Numbers 1, 2, 3, ... 2014 are written on a board.
You are allowed to replace any two of these numbers by a new number, which is either the sum or the difference of these numbers.

Show that after 2013 times performing this operation, the only number left on the board cannot be zero.

 See The Solution Submitted by Ady TZIDON No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
 re(4): All things being equal ... (spoiler) | Comment 6 of 8 |
(In reply to re(3): All things being equal ... (spoiler) by xdog)

Your interpretation was correct and still Steve is right,-

replacing 2,3 by 5 leaves the total unchanged,while replacing
the same pair by 1( or even by -1) leaves the parity unchanged.
Try a smaller set  ,any even number replacing 2014,   e.g. 1,2,3,4,5,6 and you will find out that  after 5 operations the resultremains odd.

Edited on December 13, 2013, 2:38 am
 Posted by Ady TZIDON on 2013-12-13 02:04:50

 Search: Search body:
Forums (0)