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Number Theory/Expansions

  1. May 8, 2007 #1
    1. The problem statement, all variables and given/known data
    a)True or False :
    [tex] \underbrace{111111111.....1} [/tex] is a prime number .
    [tex]91 times [/tex]

    b) Find n such that -:
    [tex] 2\times 2^2+ 3\times2^3+4\times2^4+ \cdots + n\times2^n=2^{n+10} [/tex]

    2. Relevant equations

    3. The attempt at a solution

    I have no idea about a). The number is not divisible by 3,7,11...but i cant go on dividing all the way like this. How do i resolve this into prime factors(if possible) ? Do i use binomial theorem and how ?

    About b), LHS is an AGP . I tried taking the [tex]\ n\times2^n [/tex] to RHS and then dividing by 2^2...but that doesnt seem to help ?
  2. jcsd
  3. May 10, 2007 #2
    any suggestions ?
  4. May 10, 2007 #3


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    Here are some hints...
    a) 91 is composite
    b) Perhaps you can start by taking partial sums and seeing what happens...
  5. May 14, 2007 #4
    Umm i dont understand...how does 91 being composite influence divisibility ?
    Could you plz explain in abit more detail
  6. May 14, 2007 #5


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    one reason that it helps is because 91=7x13 means that you can chop the number into segments like
    [tex]\underbrace{\underbrace{1111111}_{7\; times}\underbrace{1111111}_{7\; times}\ldots\underbrace{1111111}_{7\; times}}_{13\; times}[/tex]

    and if 1111111 is not a prime then you are done. Otherwise more work needed.
  7. May 14, 2007 #6

    Gib Z

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    [tex]11 + 1100 + 11 0000 + 11 000000 .......[/tex]

    O god damn it, 91's an odd number nvm me didn't notice
    Last edited: May 14, 2007
  8. May 14, 2007 #7


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  9. May 23, 2007 #8
    For problem (b) you might want to consider the sum to n terms of
  10. May 23, 2007 #9
    If i am not mistaken (b)'s answer is [tex]2^9+1[/tex]
    but you need to do the workings yourself..
  11. May 25, 2007 #10
    How did you get that ?

    Yeah 513 was surely an option...
  12. May 25, 2007 #11

    Gib Z

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    1111111 = 239 x 4649
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