You’re asked to solve a surprising numerical riddle:

What is the 100th digit after the decimal point in the expansion of:

\[ (1 + \sqrt{2})^{3000}\;? \]

Step 1: Use the Conjugate Trick

Let: \[ A = (1 + \sqrt{2})^{3000}, \quad B = (1 - \sqrt{2})^{3000} \]

Then:

\[ A + B \in \mathbb{Z} \]

Why? Because it’s a symmetric expression formed from a binomial expansion with irrational terms canceling in pairs.

Also:

\(B = (1 - \sqrt{2})^{3000} \approx \text{a very small number}\), since \((1 - \sqrt{2}) \approx -0.414\), and a small number raised to a high even power approaches 0.

Step 2: Integer Minus Tiny Fraction

So we know:

\[ A = \underbrace{(A + B)}_{\text{integer}} - B \]

This means \(A\) is just less than an integer, because you’re subtracting a tiny positive number from a whole number.

Since \(A\) is slightly less than an integer, its decimal expansion is:

\[ A = \text{(integer)}.9999999999\ldots \]

This continues until the decimal catches up with the tiny subtraction \(B\).

In this case, since \(B \approx (0.414)^{3000} \ll 10^{-100}\), the first at least 100 digits after the decimal are all 9.

The 100th digit after the decimal point in \( (1 + \sqrt{2})^{3000} \) is:

\[ \boxed{9} \]