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Math Help - square root problem

  1. #1
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    Zewail University - Cairo - Egypt
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    square root problem

    i have a problem in square root
    example :
    sqrt of 180 =1 √180, 2 √45, 3 √20 or 6 √5.
    but i want to calculate the maximum factor next to root that i want to calculate
    6 √5. becz 6 > 3 > 2 > 1

    i started with prime factors
    2,3,5
    and factorising 180
    =
    2
    3
    3
    2
    5

    then 180=180/5=36
    if 36 has a square root
    so it is the biggest factor
    but i failed in some examples like
    2197
    the result should be 13 * root 13


    but with my method it will be wrong
    can anyone help me
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  2. #2
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    I always look for square factors. It doesn't matter what the square factor is, because you can always simplify further later if needbe.

    \sqrt{180} = \sqrt{18\cdot 10}

     = \sqrt{2\cdot 9 \cdot 2 \cdot 5}

     = \sqrt{2^2\cdot 3^2 \cdot 5}

     = \sqrt{2^2\cdot 3^2}\sqrt{5}

     = 2\cdot 3 \sqrt{5}

     = 6\sqrt{5}.


    For the second

    \sqrt{2197} = \sqrt{13^3}

     = \sqrt{13^2}\sqrt{13}

     = 13\sqrt{13}.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    I always look for square factors. It doesn't matter what the square factor is, because you can always simplify further later if needbe.

    \sqrt{180} = \sqrt{18\cdot 10}

     = \sqrt{2\cdot 9 \cdot 2 \cdot 5}

     = \sqrt{2^2\cdot 3^2 \cdot 5}

     = \sqrt{2^2\cdot 3^2}\sqrt{5}

     = 2\cdot 3 \sqrt{5}

     = 6\sqrt{5}.


    For the second

    \sqrt{2197} = \sqrt{13^3}

     = \sqrt{13^2}\sqrt{13}

     = 13\sqrt{13}.
    thx but how can i factorize the numbers
    for example for number like
    999999999999999997
    how can i know its factors
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  4. #4
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    Divisibility tests are always a good start...
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  5. #5
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    start doing divisibilty tests and eventually dividing..
    like:
    Let abcde be a 5 digit no..

    2 :if e belongs to 0,2,4,6,8 i.e. if the no. is even, divide by 2..
    repeat if neccecary..
    3: if a+b+c+d+e =3 k .. i.e. if the sum of the digits is a multiple of 3,, then divide by 3..
    repeat if nec.
    5: if last is either 5/0

    the proofs of 7,11,13,17,19.. are non existent or are too complex to remember..
    but u must try them out for huge no.s..

    Keep dividing till u reach a prime no.
    then every time u divided,the no. is a factor.
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  6. #6
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    In fact, since you are trying to find squared factors in order to be able to reduce the square root, you only need to look at 4, 25, 36, 49, 64, 81, 100, 121, 144, 169, .... While the "divisibility tests" umanagora suggests help, you can always use a calculator to see if they divide evenly.
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