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Math Help - Discrete Factorial + Composite numbers.

  1. #1
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    Discrete Factorial + Composite numbers.

    Ok, my question is: prove that all of the following numbers are composite: 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1002

    Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?

    Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.

    Help is much appreciated!
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  2. #2
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    Quote Originally Posted by intervade View Post
    Ok, my question is: prove that all of the following numbers are composite: 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1002

    Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?

    Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.

    Help is much appreciated!

    We know that if a number divides x and y then it divides x + y and x - y.
    Well, for 2\le k \le 1002 , check that  k\,\,\,and\,\,\,1000! have at least one non-trivial divisor.

    Tonio
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  3. #3
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    Quote Originally Posted by intervade View Post
    Ok, my question is: prove that all of the following numbers are composite: 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1002

    Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?

    Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.

    Help is much appreciated!
    Not sure of a formal way to prove this, but I'll try and get my point across in a clear(ish) way.

    The number n! can be written as 1 \times 2 \times 3 \times ... \times (n-1) \times n.

    So your 1000! = 1 \times 2 \times ... \times 999 \times 1000, let's call this equation *.

    Now first we'll deal with 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1000.

    For 1000! + 2, from our definition * we know that 2 is a factor, therefore 1000! + 2 is a composite number. Use the same logic for 3-1000.

    1000! + 1001, this is the same as 1000! + (7 \times 11 \times 13). We know from * that 7, 11 and 13 are factors of 1000!, therefore 1000! + 1001 is a composite number.

    Repeat this technique for 1000! + 1002

    Not the most well written proof but hope it makes sense.
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  4. #4
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    Quote Originally Posted by intervade View Post
    Ok, my question is: prove that all of the following numbers are composite: 1000! +2, 1000! +3, 1000+4, ....., 1000! + 1002
    It should be obvious that n!, for n> 2, is composite, just from the definition. In fact, n! is divisible by every integer from n down.
    And 1000!+ 2 is divisible by 2, 1000!+ 3 is divisible by 3, 1000!+ 4 is divisible by 4, etc. Once we get to 10001 we have to be a bit more careful: 1001 is 11*91 so 1000!+ 1001 is divisible by 11, and 1002 is divisible by 2 so 1000!+ 1002 is divisible by 2.

    Now, I assume I have to start with proving that 1000! is composite or not and then prove the other numbers being added to the factorial?

    Also, our class definition of composite is: A positive integer a is called composite provided there is an integer b such that 1< b < a and b|a.

    Help is much appreciated!
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  5. #5
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    For all k with 1 \leq k \leq 1000 we have k|1000!

    And we can write: n(k) := (1000!+k) = k (\frac{1000!}{k}+ 1).
    Hence n(k) is composite.

    case: n(0) is trivial.

    We're left with n(1001), n(1002).

    Just like HallsofIvy said: Look at prime-factorisation of 1001, 1002. For all nontrivial divisors d_1|1001, d_2|1002 we can write: n(1001) = d_1(\frac{1000!}{d_1}+ \frac{1001}{d_1}) and n(1002) = d_2(\frac{1000!}{d_2}+ \frac{1002}{d_2}).

    This is because d_1,d_2 < 1000 hence d_1,d_2|1000!.
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