What digit is immediately on the right of the decimal point in $\displaystyle (\sqrt{3}+\sqrt{2})^{2002}$?

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- Aug 19th 2011, 03:59 PMalexmahoneWhat digit is immediately on the right of the decimal point?
What digit is immediately on the right of the decimal point in $\displaystyle (\sqrt{3}+\sqrt{2})^{2002}$?

- Aug 19th 2011, 05:11 PMSammySRe: What digit is immediately on the right of the decimal point?
What have you tried?

Where are you stuck? - Aug 19th 2011, 05:14 PMalexmahoneRe: What digit is immediately on the right of the decimal point?
- Aug 20th 2011, 12:09 AMmeleseIt looks right to me, but I may be wrong...
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 (^).