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

**alexmahone** · August 19th 2011, 03:59 PM

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

**SammyS** · August 19th 2011, 05:11 PM

What have you tried?

Where are you stuck?

**alexmahone** · August 19th 2011, 05:14 PM

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

**melese** · August 20th 2011, 12:09 AM

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 (^).

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

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

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

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