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Thread: 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 $\displaystyle (\sqrt{3}+\sqrt{2})^{2002}$?
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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 $\displaystyle (\sqrt{3}+\sqrt{2})^{2002} \times (\sqrt{3}-\sqrt{2})^{2002}=1$.

    Also, $\displaystyle (\sqrt{3}-\sqrt{2})^{2002}<0.1$.
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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 $\displaystyle (\sqrt{3}+\sqrt{2})^{2002}$?
    Write (*)$\displaystyle N=(\sqrt3+\sqrt2)^{2002}+(\sqrt3-\sqrt2)^{2002}$. Using the Binomial Theorem you can show that $\displaystyle N$ is an integer. (The terms with odd powers cancel each other, while those with even powers are integers.)

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

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