Results 1 to 5 of 5

Math Help - Ten Roots

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    10

    Question Ten Roots

    The problem is: Find the smallest integer greater than 1 that has a square root, a cube root, a fourth root, a fifth root, a sixth root, a seventh root, an eighth root, a ninth root, and a tenth root, all perfect.

    I have been working at this problem and the only progress I've made is that I know the interger has to end in a 0, 1, 5, or 6. And the integer is BIG, greater than 11^10.

    I know that n^2= m^3=x^4 and so on but I don't know if this is useful

    Will someone please help!! I'm so stuck
    Last edited by math61688; April 23rd 2010 at 07:26 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1576
    Awards
    1
    Quote Originally Posted by math61688 View Post
    The problem is: Find the smallest integer greater than 1 that has a square root, a cube root, a fourth root, a fifth root, a sixth root, a seventh root, an eighth root, a ninth root, and a tenth root, all perfect.
    The number 2^{2520} works.
    Do you know why?
    Is this the smallest? (use the unique factorization theorem)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Mar 2010
    Posts
    10
    No I don't know why. That number will work for all?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,610
    Thanks
    1576
    Awards
    1
    Quote Originally Posted by math61688 View Post
    No I don't know why. That number will work for all?
    Think fractional exponents: \sqrt[n]{{2^K }} = 2^{\frac{K}{n}}
    Now if 2^K has a perfect sixth root then 6|K, 6 divides K.
    Do you know/understand the unique factorization theorem?
    2520 is the smallest positive integer divisible by each of ~2,~3,\cdots,~9,~\&~10.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Mar 2010
    Posts
    10

    Smile

    Ok, yes I understand that. It's based on the LCM. Thank you!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. roots
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: December 24th 2010, 02:50 AM
  2. Roots & Imaginary Roots
    Posted in the Math Topics Forum
    Replies: 4
    Last Post: October 4th 2009, 09:24 AM
  3. roots
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 29th 2009, 05:27 AM
  4. Roots ><
    Posted in the Algebra Forum
    Replies: 3
    Last Post: October 7th 2008, 08:49 AM
  5. Given 2 imaginary roots find other 2 roots
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 26th 2008, 09:24 PM

Search Tags


/mathhelpforum @mathhelpforum