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Math Help - What digit is immediately on the right of the decimal point?

  1. #1
    MHF Contributor alexmahone's Avatar
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    What digit is immediately on the right of the decimal point?

    What digit is immediately on the right of the decimal point in (\sqrt{3}+\sqrt{2})^{2002}?
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  2. #2
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    Re: What digit is immediately on the right of the decimal point?

    What have you tried?

    Where are you stuck?
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  3. #3
    MHF Contributor alexmahone's Avatar
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    Re: What digit is immediately on the right of the decimal point?

    Quote Originally Posted by SammyS View Post
    What have you tried?

    Where are you stuck?
    I noticed that (\sqrt{3}+\sqrt{2})^{2002} \times (\sqrt{3}-\sqrt{2})^{2002}=1.

    Also, (\sqrt{3}-\sqrt{2})^{2002}<0.1.
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  4. #4
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    It looks right to me, but I may be wrong...

    Quote Originally Posted by alexmahone View Post
    What digit is immediately on the right of the decimal point in (\sqrt{3}+\sqrt{2})^{2002}?
    Write (*) N=(\sqrt3+\sqrt2)^{2002}+(\sqrt3-\sqrt2)^{2002}. Using the Binomial Theorem you can show that N is an integer. (The terms with odd powers cancel each other, while those with even powers are integers.)

    Let \alpha=(\sqrt3-\sqrt2)^{2002}. As you've noted 0<\alpha<0.1 and therefore 1>1-\alpha>0.9.
    Rearranging (*) ,we have (^) (\sqrt3+\sqrt2)^{2002}=(N-1)+(1-\alpha), meaning that 1-\alpha is the fractional part of (\sqrt3+\sqrt2)^{2002}.

    Since 1>1-\alpha>0.9, it must start like this 0.9abcd.... So \sqrt3+\sqrt2)^{2002}=N-1.9abcd..., by (^).
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