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Math Help - Class game

  1. #1
    Member u2_wa's Avatar
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    Class game

    The natural numbers from 1 to 10 are each written on the blackboard 10 times. The student in the class than play the following game: a student deletes 2 of the numbers and instead of them writes down on the blackboard their sum decreased by 1;after that another student do the same; and so on. The game continues until only one number remains on the blackboard. The remaining number is:

    a) Less than 440
    b)451
    c)460
    d)488
    e)more than 500
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  2. #2
    Member integral's Avatar
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    It should not matter what order you add them in so I will go strait down on a row of 10x1
    (1
    2)=1+1=2
    (3
    4)=3+3=6
    (5
    6)=5+5=10
    (7
    8)=7+7=14
    (9
    10)=9+9=18
    You are really only adding every odd number to itself which goes up by four past 2.
    So the sum should equal added twice all the odd numbers from 1 to the number of numbers divided by two.


    So the equation for a 10x1 should be:

    \sum_{n=1}^{5}(2n-1)+(2n-1)=50

    So you should now have to repeat this 10 time right?

    SO 10(50)=500
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  3. #3
    MHF Contributor
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    OK OK Integral; why does this one equal 10 ?
    1 2 3
    1 2 3
    1 2 3

    The answer to the original puzzle is 451, not 500:
    original sum of the 100 numbers = 550
    since 1 is deducted 99 times, then 550 - 99 = 451

    Or you can 1st do a row 1 to 10: you'll get 46 as final number; 55 - 9
    Do other rows: end up with 46 ten times; 46*10 - 9 = 451

    Formula (n by n square):
    last number = [n^2(n - 1) + 2] / 2
    Last edited by Wilmer; March 25th 2010 at 08:36 AM. Reason: add formula
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