Puzzle #195: 1989!

This is a beautiful puzzle from Mathematical Circles (Russian Experience) by Dmitri Fomin, Sergey Genkin and Ilia Itenberg.

There numbers 1, 2, 3, ….. , 1989 are written on a blackboard. It is permitted to erase any two of them and replace them with their difference. Can this operation be used to obtain a situation where all the numbers on the blackboard are zeros?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy zeroing in on the solution!

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3 Responses to Puzzle #195: 1989!

  1. Proof by construction:

    Operation applied on 1 and 2 give 1 and 1. And then applied on 1 and 1 gives 0 and 0.

    Now for any number x,
    Operation applied on 0 and x give x and x. And then applied on the two numbers gives 0 and 0.

    Apply the operation on all the numbers and get zeroes. Done.

    Solution seems too simple though. May be I understood the question wrong?

    • Suman says:

      If you choose two numbers A and B, you erase them and put |A-B| in their place. So every time you do it, you effectively reduce one count from the board.

  2. Vishal Poddar says:

    Let the numbers be a, b, c ….
    For the final sum to be zero, all the numbers will belong to one of the two groups – group 1 or group 2 such that ( sum (group 1)) – (sum (group 2)) = 0
    For example, lets the number be 6,7,8,9 then 6 and 9 will belong to group 1 and 7 and 8 to group 2.
    Now, sum(group 1 & 2) = 2* sum (group 1) which implies sum(group 1 & 2) = even number
    Sum of (1,2…., 1989) is an odd number, therefore it is not possible to get 0 finally.

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