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Posers and Puzzles

Posers and Puzzles

  1. Standard member wolfgang59
    Infidel
    23 Jun '15 02:36
    What percentage of integers contain the digit 9 (at least once)?
  2. Standard member forkedknight
    Defend the Universe
    23 Jun '15 16:03
    Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:

    1 - (8/9)(9/10)^n

    Where 'n' is the length of the integer in digits.

    I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think
  3. Standard member forkedknight
    Defend the Universe
    23 Jun '15 21:24
    What about this?

    https://www.wolframalpha.com/input/?i=summation+from+n%3D1+to+infinity+of+%28%281-%288%2F9%29%289%2F10%29%5En%29%2F%2810%5En%29%29
  4. Standard member wolfgang59
    Infidel
    23 Jun '15 22:15
    Originally posted by forkedknight
    Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:

    1 - (8/9)(9/10)^n

    Where 'n' is the length of the integer in digits.

    I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think
    Yes.
    100% of the integers have at least one occurrence of the digit "9".

  5. Subscriber AThousandYoung
    It's about respect
    29 Jun '15 00:51
    Originally posted by forkedknight
    Assuming you don't use leading zeroes, the chances that a given integer has a '9' in it is:

    1 - (8/9)(9/10)^n

    Where 'n' is the length of the integer in digits.

    I guess you might take the limit of that as 'n' approaches infinity? But that would make it very close to "all of them" I think
    If N = 1

    1 - (8/9)(9/10) = 1 - (8/10) = 1/5

    One fifth of all the one digit integers is not 9. Fail formula.
  6. Standard member forkedknight
    Defend the Universe
    02 Jul '15 20:37 / 1 edit
    Originally posted by AThousandYoung
    If N = 1

    1 - (8/9)(9/10) = 1 - (8/10) = 1/5

    One fifth of all the one digit integers is not 9. Fail formula.
    Sorry, I guess it's an off-by-one error:

    p = 1 - (8/9)(9/10)^(n-1)
  7. 30 Jul '15 11:03
    Originally posted by wolfgang59
    Yes.
    100% of the integers have at least one occurrence of the digit "9".

    Do you say that all integers has at least one nine? Then I can give you a counterexample - 8.
  8. Standard member forkedknight
    Defend the Universe
    30 Jul '15 19:45 / 2 edits
    All, except for an infinite number of them
  9. 30 Jul '15 21:03
    But - and this is not a joke:

    Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!

    This is provable. Cantor showed us how.
  10. Standard member wolfgang59
    Infidel
    05 Aug '15 02:26
    Originally posted by FabianFnas
    But - and this is not a joke:

    Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!

    This is provable. Cantor showed us how.
    Cantor cant count.
  11. Standard member forkedknight
    Defend the Universe
    05 Aug '15 18:19
    Originally posted by FabianFnas
    But - and this is not a joke:

    Among all integers there are exactly as many numbers with a nine in them as they are numbers without a nine in them. No approximately but exactly! Not as n reach infinity, but when n *is* infinity!

    This is provable. Cantor showed us how.
    Well the the answer should be 50%, and I don't think that's true...
  12. 05 Aug '15 19:37
    Originally posted by forkedknight
    Well the the answer should be 50%, and I don't think that's true...
    The number of all integers is infinitely many.
    How much is 50% of infinity?
  13. Standard member wolfgang59
    Infidel
    05 Aug '15 20:51
    Originally posted by FabianFnas
    How much is 50% of infinity?
    A half.