Calculating decimal digits by hand for extremely large numbers

Hmm. Pretty sure that the answer is $9$. The key observation to this problem is noticing that $(\sqrt{3}-\sqrt{2})^{2016}+(\sqrt{3}+\sqrt{2})^{2016}$ is an integer.

The proof of this is expansion using Binomial Theorem. The odd powers of the square roots get canceled out.

Now we have $(\sqrt{3}+\sqrt{2})^{2016} = N - (\sqrt{3}-\sqrt{2})^{2016}$, where $N$ is a positive integer.

Now this is easy. Since $(\sqrt{3}-\sqrt{2})^{2016} < (0.4)^{2016} = (0.064)^{\frac{2016}{3}} < (0.1)^{\frac{2016}{3}} < (0.1)^{600}$, we have $(\sqrt{3}+\sqrt{2})^{2016}= (N-1)+0.99\cdots 99$, and there are at least $500$ $9$'s there. The answer is $\boxed{9}$.

Compare $(\sqrt{3}+\sqrt{2})^n$ and $(\sqrt{3}-\sqrt{2})^n$
Can show their sum is an integer when $n$ is even, and the second number is very small?