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Number Theory non zero natural numbers

  1. Feb 23, 2013 #1
    1. The problem statement, all variables and given/known data

    For all non zero natural numbers n prove that:

    1- [tex]24\mid n(n+1)(n+2)(n+3)(n+4)[/tex]

    and that :

    2- [tex]120\mid n(n+1)(n+2)(n+3)(n+4)(n+5)[/tex]



    3. The attempt at a solution

    1- For n=1 we get that 24 divides 120 so we assume that 24 divides n(n+1)(n+2)(n+3)(n+4)

    and we show that 24 divides (n+1)(n+2)(n+3)(n+4)(n+5). From the first step we have that

    24 divides (n+1)(n+2)(n+3)(n+4), I can't quite finish it off. Any help would be appreciated.
     
  2. jcsd
  3. Feb 23, 2013 #2

    micromass

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    Induction is not needed.

    Can you prove that 3 always divides 3 subsequent numbers??

    Now we have to prove that 24 always divides 5 subsequent numbers. Think about the prime factorization of 24.
     
  4. Feb 23, 2013 #3
    If 3 divides 3 subsequent numbers than 3 either divides n or n+1 or n+2
     
  5. Feb 23, 2013 #4

    micromass

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    Yes, true. But it is clear why 3 always divides 3 subsequent numbers?
     
  6. Feb 23, 2013 #5
    I'm not pretty sure but i believe it is that when 3 subsequent numbers are multiplied together they are always in the from 3k(3k+1)(3k+1) and making that 3k is always divisible by 3.
     
  7. Feb 23, 2013 #6

    micromass

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    Good. Now perhaps simplify the question in your OP by splitting 24 up in factors.
     
  8. Feb 23, 2013 #7
    Ok when splitting up 24 into factors i got that : [tex]24=2^{3}*3[/tex]

    And 3 divides n(n+1)(n+2)(n+3)(n+4) because 3 divides n(n+1)(n+2) and 2^3 divides (n+1)(n+2)(n+3)(n+4) because 2^3 divides n(n+1)(n+2)(n+3) and 3 and 8 are prime to each other so 3*2^3 divides n(n+1)(n+2)(n+3)(n+4) right?
     
    Last edited: Feb 23, 2013
  9. Feb 23, 2013 #8

    micromass

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    Yes. Go on... Do the factors individually need to divide the number?
     
  10. Feb 23, 2013 #9
    Sorry i edited my previous post while you were replying.
     
  11. Mar 6, 2013 #10
    The problem appears to be on the unambitious side. Actually, for ##n \in \mathbb{N}_1##,

    a) ## 24\text{ | } n(n+1)(n+2)(n+3)##

    and

    b) ##120 \text{ | } n(n+1)(n+2)(n+3)(n+4)##

    Think modularly...
     
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