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.
(In reply to
re(3): All things being equal ... (spoiler) by xdog)
If at the end of the series of operations, only one number is left on the board, that means that each step reduced the number of numbers remaining on the board, in fact by one. At each step, where there were two numbers before, now there are only one, so for example:
1, 2, 3, 4, 5, ...
could become
1, 5, 4, 5, ...
and the overall parity is the same.
In fact, in your second interpretation where one number is replaced by the sum and the other by the difference, parity wouldn't be preserved:
1, 2, 3, 4, 5, ...
being replaced by
1, 1, 5, 4, 5, ...
would have changed the parity.

Posted by Charlie
on 20131213 01:54:49 