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K|n!+k from k=2,3,4 .n

  1. Jul 8, 2014 #1
    1. The problem statement, all variables and given/known data

    prove by induction that K|n!+k for all integers [tex]n \ge 2 [/tex] and k = 2,3,4,......,n

    this one is different than other proofs I have had because Ive only dealt with n changing but now n and k change.

    k goes up to n, does this mean k and n have the same values?

    for instance,


    [tex]2|2! + 2 \Rightarrow 2|4[/tex]

    so p(k)

    is [tex]k|n!+k; n=k[/tex]


    Last edited: Jul 8, 2014
  2. jcsd
  3. Jul 8, 2014 #2

    Ray Vickson

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    You ask: "does this mean k and n have the same values?". No, that is not what it says and not what it means. Look at some small-n examples. If n = 3 it says that 2|3!+2 and 3|3!+3. So, k is not 3; it is any number in the set {2,3}. If n = 4 it says that all three of the statements 2|4!+2, 3|4!+3 and 4|4!+4 are true.
  4. Jul 8, 2014 #3
    ah okay

    whatever n is, then k takes on each value in the set {2,...,n}
  5. Jul 8, 2014 #4


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    The fact that if k< n the n k divides n! makes this pretty close to trivial.
  6. Jul 8, 2014 #5

    where k takes on the values of 2,3,4...n and n is an integer greater than or equal to 2

    and because for all k<n, k|n and when n=k, k|n

    so my p(k) is




    Last edited: Jul 9, 2014
  7. Jul 9, 2014 #6
    I have to make a correction. its not asking for induction. all the other problems on this problem set were induction but I confirmed with my prof this is not an induction proof question
  8. Jul 9, 2014 #7


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    Can you see a common factor in the numerator?
  9. Jul 9, 2014 #8
    k is a factor of k, clearly and k is a also a factor of n! by the given information

    [tex]2 \le k \le n[/tex] and by definition of a factorial k had to be used somewhere along the product

    [tex] \prod_{k=2}^{n} k[/tex]

    case 1

    k = 2, n > or equal 2

    [tex] \frac{2*.....*n + 2}{2}[/tex]

    case 2

    2 < k <n

    [tex]\frac{2*...*k*....*n + k}{k}[/tex]

    case 3


    [tex]\frac{2*......*k + k}{k}[/tex]
    Last edited: Jul 9, 2014
  10. Jul 9, 2014 #9


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    Those case are not necessary. WWDG's point is that if [itex]k\le n[/itex] then k is a factor of n!.
  11. Jul 9, 2014 #10
    I understand that but I am expected to show a proof. Seems a tad bit too easy to say if

    [tex] k \le n[/tex] then k is a factor of n!
  12. Jul 9, 2014 #11

    Ray Vickson

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    But that is absolutely correct, so IS a proof! Sometimes proofs are easy---after you "see" them.
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