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Math Help - lcm of k,k+1,k+2 where k≡3(mod 4)

  1. #1
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    lcm of k,k+1,k+2 where k≡3(mod 4)

    Find the least common multiple of k,k+1, and k+2 where k≡3(mod 4).

    Some ideas:
    k≡3(mod 4) => k=3+4z for some z E Z
    So k is odd, k+1 is even, and k+2 is odd.
    Also, this question asks about the lcm of THREE numbers, so I don't think the the Euclidean algorithm (which only gives the gcd or lcm of TWO numbers) will help us in this question.

    So how can we solve this?
    Thanks for any help!


    [also under discussion in math links forum]
    Last edited by kingwinner; February 12th 2010 at 01:51 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by kingwinner View Post
    Find the least common multiple of k,k+1, and k+2 where k≡3(mod 4).

    Some ideas:
    k≡3(mod 4) => k=3+4z for some z E Z
    So k is odd, k+1 is even, and k+2 is odd.
    Also, this question asks about the lcm of THREE numbers, so I don't think the the Euclidean algorithm (which only gives the gcd or lcm of TWO numbers) will help us in this question.

    So how can we solve this?
    Thanks for any help!
    Hint: \text{lcm}\left(a,b,c\right)=\text{lcm}\left(\text  {lcm}(a,b),c\right)
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  3. #3
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    Quote Originally Posted by Drexel28 View Post
    Hint: \text{lcm}\left(a,b,c\right)=\text{lcm}\left(\text  {lcm}(a,b),c\right)
    Is this always true? How can we prove it?
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  4. #4
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    Since k is odd, k+1 is even, and k+2 is odd
    then
    GCD(k,k+1,k+2)=1 => LCM(k,k+1,k+2)=k(k+1)(k+2)
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  5. #5
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    Quote Originally Posted by dooping View Post
    Since k is odd, k+1 is even, and k+2 is odd
    then
    GCD(k,k+1,k+2)=1 => LCM(k,k+1,k+2)=k(k+1)(k+2)
    How do we actually formally prove these two claims?
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