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Last but not the least (Posted on 2013-12-12) Difficulty: 2 of 5
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    
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re(4): All things being equal ... (spoiler) | Comment 5 of 8 |
(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 2013-12-13 01:54:49
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