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Math Help - Algebra help.Finding four whole numbers that agree with n

  1. #1
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    Algebra help.Finding four whole numbers that agree with n

    Hi guys this is first post on the forum as I am seeking help for this question

    The question states:

    5. Some statements have the interesting property that they describe themselves. For example:
    “This sentence contains thirty six letters”

    (i) Find a list of four whole numbers n0, n1, n2, n3 which obeys the following rules

    n0 = the number of times 0 occurs in your list
    n1 = the number of times 1 occurs in your list
    n2 = the number of times 2 occurs in your list
    n3 = the number of times 3 occurs in your list

    Explain carefully how you obtained your list.

    (ii) Find another list of four, different from your answer in part (i)
    (iii) Is it possible to form a similar list with only two numbers n0, n1 ? Explain carefully.
    (iv) Is it possible to form a similar list with only three numbers n0, n1, n2 ? Explain carefully.
    (v) In a list with eight numbers n0, n1, n2, n3, n4, n5, n6, n7 could n4, n5, n6, n7 all be zero? Explain.



    Could please help me with at least i)
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  2. #2
    Super Member ILikeSerena's Avatar
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    Re: Algebra help.Finding four whole numbers that agree with n

    Hi gazing600000!

    It's a matter of checking the different cases.

    The number of zeroes is either 0,1,2,3, or 4.

    Suppose there are 0 zeroes, then your list is (0,x,x,x).
    But this list already contains a zero so that is a contradiction.

    Suppose there are 4 zeroes, then your list is (4,0,0,0).
    But 4 zeroes don't even fit, since the first number (n0) is 4.

    And so on...

    Can you do a couple of cases?
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