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Math Help - Proof by Induction (Divisibility)

  1. #1
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    Proof by Induction (Divisibility)

    Prove by induction the following:

    10^(n+1) + 3*10^n + 5 is divisable by 9

    I got the base step correct. Here is what I had for the inductive step:

    Assume Pk is true, 9|10^(k+1) + 3*10^k + 5
    Consider Pk+1
    10^(k+2) + 3*10^(k+1) + 5
    10(10^(k+1)) + 3*10^k(10) + 5
    (10^(k+1) + 3*10^k + 5) (10*10)
    9|(10^(k+1) + 3*10^k + 5) (10*10) because 9|10^(k+1) + 3*10^k + 5

    Any help would be greatful.
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  2. #2
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    Quote Originally Posted by page929 View Post
    Prove by induction the following:

    10^(n+1) + 3*10^n + 5 is divisible by 9

    I got the base step correct. Here is what I had for the inductive step:

    Assume Pk is true, 9|10^(k+1) + 3*10^k + 5
    Consider Pk+1
    10^(k+2) + 3*10^(k+1) + 5
    10(10^(k+1)) + 3*10^k(10) + 5

    At this point you should be seperating out.... 10^{k+1}+(3)10^k+5

    to get... 10\left(10^{k+1}+(3)10^k\right)+5=9\left(10^{k+1}+  (3)10^k\right)+10^{k+1}+(3)10^k+5



    (10^(k+1) + 3*10^k + 5) (10*10)
    9|(10^(k+1) + 3*10^k + 5) (10*10) because 9|10^(k+1) + 3*10^k + 5

    Any help would be greatful.
    Then if P(k) is really true, all of that will be divisible by 9.
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  3. #3
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    You can also :
    Proof by Induction (Divisibility)-img.jpg
    Last edited by Pandevil1990; October 8th 2010 at 02:20 PM.
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