# 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 $(\sqrt{3}+\sqrt{2})^{2002}$?
• 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 $(\sqrt{3}+\sqrt{2})^{2002} \times (\sqrt{3}-\sqrt{2})^{2002}=1$.

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

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 (^).