Results 1 to 2 of 2

Thread: m!+ K

  1. #1

    Thread Starter
    New Member
    Join Date
    Apr 2007
    Posts
    8

    m!+ K

    I Have been given this question and cannot get an answer so here goes if anyone can help within the next hour would be great.

    Explore the divisors of m! + k for various m and k. Hence explain how to find a run of n consecutive integers which are composite.

    Example:

    m=5

    5! = 120  Composite
    5! + 1 = 121  Composite
    5! + 2 = 122  Composite
    5! + 3 = 123  Composite
    5! + 4 = 124  Composite
    5! + 5 = 125  Composite
    5! + 6 = 126  Composite
    5! + 7 = 127  Not Composite

    The ? are suppose to be arrows.

  2. #2
    Only Slightly Obsessive jemidiah's Avatar
    Join Date
    Apr 2002
    Posts
    2,431

    Re: m!+ K

    [The only truly interesting m are all greater than 1, so assume m>1]

    When k=0, m!+k is of course composite.

    In the example you had, you can also write 5!+3 as 5*4*3*2*1 + 3 = 3*(5*4*2*1 + 1) = 3*(5!/3 + 1) so 5!+3 is composite. In fact, you can write any of these m!+k as k*(m!/k+1). Here, 3 | 5!, so m!/k + 1 is an integer. So long as k > 1 and k | m!, you can factor m!+k like this into two numbers greater than 1, making m!+k composite.

    You should probably be able to find the n consecutive integers using the ideas above. If not, tell me what you get and I'll fill in some more.
    The time you enjoy wasting is not wasted time.
    Bertrand Russell

    <- Remember to rate posts you find helpful.

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  



Click Here to Expand Forum to Full Width