# What digit is immediately on the right of the decimal point?

• Aug 19th 2011, 03:59 PM
alexmahone
What digit is immediately on the right of the decimal point?
What digit is immediately on the right of the decimal point in $\displaystyle (\sqrt{3}+\sqrt{2})^{2002}$?
• Aug 19th 2011, 05:11 PM
SammyS
Re: What digit is immediately on the right of the decimal point?
What have you tried?

Where are you stuck?
• Aug 19th 2011, 05:14 PM
alexmahone
Re: What digit is immediately on the right of the decimal point?
Quote:

Originally Posted by SammyS
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, $\displaystyle (\sqrt{3}-\sqrt{2})^{2002}<0.1$.
• Aug 20th 2011, 12:09 AM
melese
It looks right to me, but I may be wrong...
Quote:

Originally Posted by alexmahone
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 (^).