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Proving 11 is a factor of 2^(4n+3)+3(5^n)

  1. Apr 22, 2012 #1
    1. The problem statement, all variables and given/known data

    Theorem if n is a non-negative integer then 11 is a factor of 2^(4n+3)+3(5^n)

    Basic ideas if 2^(4n+3) =11k -3(5^n) then,
    2^(4n+7) =11(16k -3(5^n))-3(5^(n+1))



    2. Relevant equations



    3. The attempt at a solution

    So I understand how they get the result 2^(4n+3) =11k -3(5^n)

    but I have no idea how to obtain the second result, namely, 2^(4n+7) =11(16k -3(5^n))-3(5^(n+1)) ????

    Im stumped I can obviously see how they have added another 2^4 but I really dont see how that produces RHS. Help!
     
  2. jcsd
  3. Apr 22, 2012 #2
    Mathematical induction, [itex]k \rightarrow k+1[/itex]
     
  4. Apr 22, 2012 #3
    Ok but when I do my basis step for n=1...

    2^(4+3) +15 = 11k

    so 2^7 +15 =11k

    so 143= 11(13) which is true

    so if 2^(4n+3)=11k -3(5^n)...

    so 11k = 2^(4n+3) +3(5^n)

    now check its true for n+1

    so 2^(4(n+1)+3) =11K-3(5^(n+1))

    so 2^(4n+7) = 11k-3(5^(n+1))

    so subing in for 11k from above I have

    2^(4n+7) = (2^(4n+3)+3(5^n))-3(5^(n+1))'

    but I need to show

    2^(4n+7)=11(16k-3(5^n))-3(5(n+1))

    ???

    I have not really done any mathematical induction before I just watched a few videos, so I have probably done something wrong. Help!

    so 1
     
  5. Apr 22, 2012 #4

    Curious3141

    User Avatar
    Homework Helper

    This shouldn't be under Calculus. :wink:

    Here's a hint:

    [tex]2^{4(n+1) + 3} + 3.5^{n+1} = 16.2^{4n + 3} + 5.(3.5^n)[/tex]

    and 16 = 5 + 11. Regroup those factors! :biggrin: Try and manipulate it into something like:

    [tex]p.(2^{4n + 3} + 3.5^n) + 11q[/tex]

    where p and q are integers or integral expressions that I'll leave you to determine.

    Once you've assumed the inductive hypothesis that [itex](2^{4n + 3} + 3.5^n)[/itex] is a multiple of 11, wouldn't that whole expression magically become a multiple of 11 too?

    Then all you need to do is test it for n=1.:smile:
     
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