Results 1 to 5 of 5

Thread: Discrete Factorial + Composite numbers.

  1. #1
    Newbie
    Joined
    Jan 2010
    Posts
    19

    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!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    3
    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 $\displaystyle 2\le k \le 1002$ , check that $\displaystyle k\,\,\,and\,\,\,1000!$ have at least one non-trivial divisor.

    Tonio
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member craig's Avatar
    Joined
    Apr 2008
    Posts
    748
    Thanks
    1
    Awards
    1
    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 $\displaystyle n!$ can be written as $\displaystyle 1 \times 2 \times 3 \times ... \times (n-1) \times n$.

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

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

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

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

    Repeat this technique for $\displaystyle 1000! + 1002$

    Not the most well written proof but hope it makes sense.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    19,798
    Thanks
    3035
    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!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Dinkydoe's Avatar
    Joined
    Dec 2009
    Posts
    411
    For all $\displaystyle k$ with $\displaystyle 1 \leq k \leq 1000$ we have $\displaystyle k|1000!$

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

    case: $\displaystyle n(0)$ is trivial.

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

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

    This is because $\displaystyle d_1,d_2 < 1000$ hence $\displaystyle d_1,d_2|1000!$.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Composite numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Mar 6th 2010, 11:44 PM
  2. composite numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Sep 10th 2009, 12:52 PM
  3. Composite numbers
    Posted in the Number Theory Forum
    Replies: 3
    Last Post: Jan 5th 2009, 10:19 AM
  4. [SOLVED] factorial for big numbers
    Posted in the Algebra Forum
    Replies: 13
    Last Post: Mar 14th 2007, 12:55 AM
  5. Composite Numbers
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: Feb 19th 2007, 07:41 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum