Hundredth decimal digit of a huge power of 1 plus root 2

Consider the number $(1+\sqrt2)^{3000}$. Find the digit sitting in the 100th place after its decimal point.

$9$. The sum $(1+\sqrt2)^{3000}+(1-\sqrt2)^{3000}$ is an integer, since the odd powers of $\sqrt2$ cancel in the binomial expansions. As $0<(\sqrt2-1)^{3000}<10^{-1000}$, the original number equals an integer minus a tiny positive amount, so its first several hundred decimal digits are all $9$.