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

1. What digit is immediately on the right of the decimal point?

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

2. Re: What digit is immediately on the right of the decimal point?

What have you tried?

Where are you stuck?

3. Re: What digit is immediately on the right of the decimal point?

Originally Posted by SammyS
What have you tried?

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

Also, $(\sqrt{3}-\sqrt{2})^{2002}<0.1$.

4. It looks right to me, but I may be wrong...

Originally Posted by alexmahone
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 (^).