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Proving this

  1. Sep 13, 2004 #1
    is there a way to prove that n consecutive integers is always divisible by n!?

    thanks in advance
     
  2. jcsd
  3. Sep 13, 2004 #2

    Gokul43201

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    I think you have worded this question incorrectly. Going strictly by your wording, the answer would be : No...in fact, there are no two consecutive numbers that are divisible by any given number ( >1).

    I think I know what you might be intending to ask, but I'll let you pose the question correctly, first.
     
  4. Sep 13, 2004 #3
    sorry, but i dont know what your saying, where did two come from?
     
  5. Sep 13, 2004 #4

    Gokul43201

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    If no two consecutive integers are divisible by any integer ( >1), how can you find n consecutive integers ?
     
  6. Sep 13, 2004 #5

    Alkatran

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    starting equations:
    n mod y = 0
    (n+1) mod y = 0

    work:
    ((n mod y) + (1 mod y)) mod y = 0
    (n mod y) + 1 = 0
    (n mod y) + 1 = n mod y
    1 = (n mod y) - (n mod y)
    1 = 0

    conclusion:
    FALSE

    No two consecutive integers are divisible by the same factor.

    That's what he's talking about. (THIS IS NOT YOUR ANSWER)
     
  7. Sep 13, 2004 #6
    Easier: let n=3. The statement is that 3 consequtive integers is divisible by 6. Well, 1,2,3 provides a counterexample.

    Wait. I'm not sure what you mean now. Do you mean the SUM of n consequtive integers is divisible by n!?

    If that's what you meant, then 2, 3, 4 has sum 9 and is not divisible by 6(=3!).
     
  8. Sep 13, 2004 #7

    Gokul43201

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    The more common question is : Prove that you can always find n consecutive composite numbers.

    Of course, these are be (n+1)!+2, (n+1)!+3, (n+1)!+4, ...(n+1)!+(n+1), and the proof is quite self-evident.
     
  9. Sep 13, 2004 #8

    Gokul43201

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    Just to make this complete (not to nitpick)

    starting equations:
    n mod y = 0
    (n+1) mod y = 0

    work:
    ((n mod y) + (1 mod y)) mod y = 0
    (n mod y) + 1 = 0 {y <> +/- 1}
    (n mod y) + 1 = n mod y
    1 = (n mod y) - (n mod y)
    1 = 0

    conclusion:
    FALSE

    No two consecutive integers are divisible by the same factor (>1).
     
  10. Sep 13, 2004 #9

    Alkatran

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    I was thinking about that when I wrote it. I got lost on a tangent of what the remainder of mod 0 was... undefined, bah.

    Gimme a break, it's the second time I try to prove something using modulo.

    I have a question though, is:
    (x op y) mod z == ((x mod z) op (y mod z)) mod z
    true? (op being +, -, *, / , ^, etc) is it true for functions? (sin, f())
     
  11. Sep 13, 2004 #10

    Gokul43201

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    Sorry 'bout that.

    To answer your question : In general, NO.
    It is true, however for +, - and *.

    I'm not sure how your question applies to functions like sin(x), when you are asking about binary operators.
     
  12. Sep 13, 2004 #11

    Alkatran

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    So, could I say:

    If x op y = y op x then (x op y) mod z = ((x mod z) op (y mod z)) mod z

    Alright, the sin() thing won't work, I was off on a tangent. What about binary operations on the numbers? (XOR, OR, AND)
     
  13. Sep 13, 2004 #12

    Gokul43201

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    To clarify, when I said 'binary operator', I meant an operator defined on 2 numbers (such as addition, subtraction, exponentiation, etc), not operators defined on the binary representation of numbers.

    I'm pretty sure that rule doesn't work for AND, OR, XOR, etc...but I haven't verified this.

    "If x op y = y op x then (x op y) mod z = ((x mod z) op (y mod z)) mod z"

    In general, this is not necessarily true.
     
    Last edited: Sep 13, 2004
  14. Sep 13, 2004 #13

    Alkatran

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    I understand what you meant by binary operators. It just made me think of the logical operations. (computer programmer here...)

    Do you have an example or shall I find one myself?
    Got it:
    5 and 7 = 5
    5 mod 3 = 2
    (5 and 7) mod 3 = 2
    ((5 mod 3) and (7 mod 3)) mod 3 = (2 and 1) mod 3 = 0 mod 3 = 0
    0 <> 2
     
    Last edited: Sep 13, 2004
  15. Sep 13, 2004 #14
    i know there's a theorem similar to what decibel asked about in lozansky's "winning solutions". i can't remember exactly how it went though, but it definitely had something about n consecutive integers & n!
     
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