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(n-1)! is divisible by n

  1. Apr 4, 2008 #1

    xax

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    I need to prove this for any n natural, n>= 5, n not prime.
     
  2. jcsd
  3. Apr 4, 2008 #2
    Think a bit about the prime factors of n... are they smaller than n? Then think of what (n-1)! means.
     
  4. Apr 4, 2008 #3

    HallsofIvy

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    Dodo's hint is excellent- but the crucial point is whether the factors of n are less than n-1, not jus n itself!
     
  5. Apr 4, 2008 #4

    mathman

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    If n is not prime, then n=j*k, where j and k are >1, therefore both <n. As a result both are factors in (n-1)!, so n divides (n-1)!.
     
  6. Apr 4, 2008 #5
    Almost there, one last thing to consider is the squares of prime numbers. What happens when n=9=3*3 for example? This is where the n>=5 comes in.
     
  7. Apr 5, 2008 #6

    xax

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    You all have good points and I thought of all of them and the reason I've posted this is because of the square numbers as rodigee said. Well I think if n = k*k then k is one of the numbers between 1 and n-1. since n=k*k then k divides (n-k) and this is smaller the n-1 which means n divides (n-1)!.
     
  8. Apr 5, 2008 #7
    Sorry, xax but I don't really understand your idea heh. This is how I see it:

    [tex]n > 4[/tex] implies that

    [tex]\sqrt{n} > 2[/tex] so we also have that [tex]\sqrt{n}\sqrt{n} > 2\sqrt{n}[/tex] or simply [tex]n > 2\sqrt{n}[/tex], meaning that [tex]2\sqrt{n}[/tex] and [tex]\sqrt{n}[/tex] both show up in [tex](n-1)![/tex] so we're good to go.
     
  9. Apr 5, 2008 #8

    xax

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    You make perfect sense rodigee and this is a nicer demonstation than mine. Thank you.
     
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