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Math Help - A number riddle

  1. #1
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    A number riddle

    I created a riddle and supplied an answer, I happen to think it is interesting.
    As one counts from 1 to 1000000000 how many times does one meet the number nine? An example: from 1 to 100 is 20 because of: 9, 19, 29, 39, 49, 59, 69, 79, 89, 90, 91, 92, 93, 94, 95, 96, 97, 98, 99 and 99 counts as two! So how many times do I meet the number 9 from 1 to 1000000000? You can always count or find a pattern.

    Okay the solution is this:
    on the interval:
    [1,10]=1
    [1,100]=20
    [1,1000]=300
    [1,10000]=4000
    [1,100000]=50000
    See the pattern! We can in fact prove from [1,10^n] we meet the number nine n*10^(n-1) times. But the proof is ommitted here.

    Tell me what you think of it?

    Something interesting happens when we count to 10 trillion the number of times we meet 9 is also 10 trillion! Futhermore when the numbers start getting larger than 10 trillion the number of times we meet the nine exceedes the number we count until!
    Last edited by ThePerfectHacker; December 10th 2005 at 06:10 PM.
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  2. #2
    Member
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    Hi:

    And interesting it is indeed!
    For a 1 digit number, we see 9 once.

    2 digits: __ __, __ __ , __ __ The 9 can show up in the 1st, 2nd, or both places. If it only appears in one position, then the other position can be filler by any of 9 digits. And it can show in both places only one way, i.e., 99, where we see the 9 twice. The total "n(2)" = (2c1)X9^1 + 2(2c2)X9^0 = 20.

    Employing the same logic gives,
    n(3) = (3c1)X9^2 + 2(3c2)X9^1 + 3(3c3)x9^0 = 300.
    n(4) = (4c1)X9^3 + 2(4c2)X9^2 + 3(4c3)x9^1 + 4(4c4)x9^0 = 4000
    .
    .
    .
    n(k)= (kc1)x9^(k-1) + 2(kc2)x9^(k-2) +...+k(kck)x9^(k-k) = kX10^(k-1).

    Hence, as the number of appearances in {0,1,2,...,10^9} is the same as those in {0,1,2,...,999999999}, the solution of your specific question is,
    n(9) = 9X10^8 or 900,000,000.

    With regard to proof, it should be fairly short business to demonstrate that
    n(k)= kX10^(k-1) implies n(k+1)=(k+1)X10^[(k-1)+1] for k in {1,2,3,...}, making induction a reasonably "clean" course of action. ...best laid plans...

    Regards,

    Rich B.
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