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.
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 $0.999.....$