Numbers 1, 2, 3, ... 2014 are written on
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.
(In reply to re(3): All things being equal ... (spoiler)
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