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09/03/2017 at 09:39
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(1974 Kiew Math Olympiad)

Numbers 1, 2, 3, ..., 1974 are written on the board. You are allowed to replace any two of these numbers by one number, which is either the sum or the difference of these two numbers. Show that after 1973 times performing this operations, the only number left on the board cannot be 0.


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    hghfghfgh 26/03/2017 at 20:16

    At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

    + if a odd and b even, or a even and b odd then a + b or a - b is still odd

    + if a and b are both even then a + b or a -b is still even

    + If a and  are both odd then a + b or a - b is even

    So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers  (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).haha

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    Faded 19/01/2018 at 14:52

    At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

    + if a odd and b even, or a even and b odd then a + b or a - b is still odd

    + if a and b are both even then a + b or a -b is still even

    + If a and  are both odd then a + b or a - b is even

    So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers  (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even). [haha]

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    An Duong 10/03/2017 at 14:14

    At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

    + if a odd and b even, or a even and b odd then a + b or a - b is still odd

    + if a and b are both even then a + b or a -b is still even

    + If a and  are both odd then a + b or a - b is even

    So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers  (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).

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    xicor 24/08/2017 at 10:39

    coppy là gian nghe

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    Nguyễn Tiến Dũng 29/08/2017 at 13:03


     

    At the begining, there are 1974/2=987 odd numbers. When we replace two numbers a and b by a number either a + b or a - b, we see that:

    + if a odd and b even, or a even and b odd then a + b or a - b is still odd

    + if a and b are both even then a + b or a -b is still even

    + If a and  are both odd then a + b or a - b is even

    So the number of odd after each such replacement operation is stay the same or deacresed by two. At the begining, there is 987 odd numbers  (987 is odd) and the odd number left must be odd too. So at the final, only one number left, it must be odd number, it is not 0. (because 0 is even).

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    Help you solve math 28/08/2017 at 20:46

    We are copy

  • ...
    Help you solve math 13/08/2017 at 08:52

    Copy nhau à



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