$1000$th decimal digit of $(8+\sqrt{63})^{2012}$ 
Find the digit at the $1000$th position at the right of the decimal point of the number $(8+\sqrt{63})^{2012}$

I took this problem from a Mexican Math Olympiad called Galois-Noether. It's the last problem here. The options are $1$, $3$, $7$ and $9$ and the correct one is $9$.
I don't know how to solve this. I know that if we write $(8+\sqrt{63})^{2012}$ as $n.d_1d_2\ldots d_{1000}\ldots$, where $n$ is the integer part of the number and $d_i=0,1,\ldots,9$, then we'll get
$$10^{1000}(8+\sqrt{63})^{2012}=nd_1d_2\ldots d_{1000}.d_{1001}\ldots$$
So all we have to do is to take the integer part of $10^{1000}(8+\sqrt{63})^{2012}$ and then take it $\;\text{mod}\,10$. But, unless you use a computer, this seems too complicated.
One thing I think that'll might help is the fact that $\sqrt{63}$ is ''close'' to $\sqrt{64}=8$. But I can't see how to use this.
 A: 1) $(8-\sqrt{63})^{2012}=N.0000000...00ABC...$
2) $(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}=A \in Z$
Then 
3) $(8+\sqrt{63})^{2012}=K.9999999...99DCE...$
A: Hints: 
(i) $(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$ is an integer. One way to prove this is to use the binomial theorem.
(ii) $(8-\sqrt{63})^{2012}$ is positive and quite a bit less than $10^{-1000}$. (The fact that $\sqrt{63}$ is close to $8$ is useful for this part).
A: As André Nicolas has kindly pointed out, 
$$(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$$ is equal to an integral value. Also, $(8-\sqrt{63})^{2012}$ is equal to a fractional value. Thus, we can write:
$$I + f = (8+\sqrt{63})^{2012}$$
$$f` = (8-\sqrt{63})^{2012}$$
Where I means integer while f means fractional part.
Addding the two, we get:
$$I + f + f` = (8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$$
Now, as both f and $f`$ are between 0 and 1, their sum will be between 0 and 2. However, we know that $I + f + f`$ is equal to a integral value and thus $f + f`$ can only equal to 1. Hence:
$$f + f` = 1$$
$$f = 1 - f`$$
As the value of $f`$ is quite small, $1-f`$ will give a number very close to 1, i.e $0.999.....$
