1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Common multiple of positive integers

  1. Apr 24, 2008 #1
    Just have this question im having trouble with

    The least common multiple of positive integers a, b, c and d is equal to a + b + c + d.
    Prove that abcd is divisible by at least one of 3 and 5.

    Thanks
     
  2. jcsd
  3. Apr 24, 2008 #2
    does anyone have an answer?
     
  4. Apr 25, 2008 #3
    anyone out there???
     
  5. May 5, 2008 #4
    man I'll give it a try but i'm not math pro so bear with me.

    (1) lcm= a+b+c+d = x1a = x2b = x3c = x4d for x1-x4 are some integer.

    raise to exponent 4 on the left side will yield : (a+b+c+d)^4=(x1a)(x2b)(x3c)(x4d)
    rewrite it to : (a+b+c+d)^4 = x1x2x3x4(abcd)
    divide by x1x2x3x4 : (a+b+c+d)^4/(x1x2x3x4)=abcd
    divide both side by 15 which is 3 and 5 : (a+b+c+d)^4/(15x1x2x3x4)=abcd/15
    now we want the left side to equal to some integer say k, let have k+1 for simplicity
    k=1=(a+b+c+d)^4/(15x1x2x3x4)
    rearange : x1x2x3x4=(a+b+c+d)^4/(15)
    If we take a+b+c+d = 15, then x1x2x3x4=15^3 = 3375 we can expand this number to get some random x1-4. Though the equation is valid but you have to also satisfy the first requirement. Sorry, coudln't help u. Hope this might give some idea.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Common multiple of positive integers
  1. Common Multiples Proof (Replies: 10)

  2. Integer Multiple of A (Replies: 7)

Loading...