Page 2 of 2 FirstFirst 12
Results 16 to 19 of 19

Math Help - 26th digit

  1. #16
    Newbie
    Joined
    May 2006
    Posts
    13
    can we use the Divisibility rule for 13?
    http://en.wikipedia.org/wiki/Divisibility_rule
    tnx
    Follow Math Help Forum on Facebook and Google+

  2. #17
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by malaygoel
    I would be happy if you post here the proof of the statement in the quoted work.But the question still remains:Is there any criteria by which we can say that " \frac{10^k - 1}{9} is prime or composite depends on the value of k"?
    Similarly, what can you say about the numbers 10^k + 1(k\geq4).
    Keep Smiling
    Malay
    Assume,
    \frac{10^k-1}{9} is a prime and k is not.
    Then,
    \frac{(10^a)^b-1}{9} where a,b are proper non-trivial factors of k.
    Using the identity,
    x^n-y^n=(x-y)(x^{n-1}y+...+xy^{n-1})
    Thus,
    \frac{(10^a-1)((10^a)^{b-1}+(10^a)^{b-2}+...+1)}{9}
    Using, this identity again,
    \frac{(10-1)(10^{a-1}+...+1)((10^a)^{b-1}+...+1)}{9}
    You are left with,
    (10^{a-1}+...+1)((10^a)^{b-1}+...+1)
    Since, a,b\not = 1 These two factors are not trival. Thus,
    \frac{10^k-1}{9} cannot be prime a contradiction.
    Follow Math Help Forum on Facebook and Google+

  3. #18
    Super Member malaygoel's Avatar
    Joined
    May 2006
    From
    India
    Posts
    648
    Quote Originally Posted by ThePerfectHacker
    Assume,
    \frac{10^k-1}{9} is a prime and k is not.
    Then,
    \frac{(10^a)^b-1}{9} where a,b are proper non-trivial factors of k.
    Using the identity,
    x^n-y^n=(x-y)(x^{n-1}y+...+xy^{n-1})
    Thus,
    \frac{(10^a-1)((10^a)^{b-1}+(10^a)^{b-2}+...+1)}{9}
    Using, this identity again,
    \frac{(10-1)(10^{a-1}+...+1)((10^a)^{b-1}+...+1)}{9}
    You are left with,
    (10^{a-1}+...+1)((10^a)^{b-1}+...+1)
    Since, a,b\not = 1 These two factors are not trival. Thus,
    \frac{10^k-1}{9} cannot be prime a contradiction.
    Great!
    You proved if k is composite
    \frac{10^k-1}{9} is composite.
    What if k is prime?

    KeepSmiling
    Malay
    Follow Math Help Forum on Facebook and Google+

  4. #19
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by malaygoel
    What if k is prime?
    Unknown.
    Sometimes prime and sometimes is not.
    If it were, then we would have the very first case of a prime producing phunction!
    Follow Math Help Forum on Facebook and Google+

Page 2 of 2 FirstFirst 12

Similar Math Help Forum Discussions

  1. The last digit of 7^7^7
    Posted in the Number Theory Forum
    Replies: 7
    Last Post: November 5th 2011, 05:59 AM
  2. Digit sum & digit product of number x
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 19th 2011, 08:07 AM
  3. Last digit
    Posted in the Algebra Forum
    Replies: 2
    Last Post: June 28th 2009, 02:37 AM
  4. digit nos
    Posted in the Statistics Forum
    Replies: 1
    Last Post: March 28th 2009, 08:06 AM
  5. decimal digit as final digit
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: February 25th 2008, 07:18 PM

Search Tags


/mathhelpforum @mathhelpforum