What is the 100th digit after the decimal point?

For (1 + √2)3000
Start
x = (1 + √2)3000

Directly computing this enormous number is unnecessary. The trick is to introduce its conjugate.

Conjugate
y = (1 − √2)3000

Since |1 − √2| ≈ 0.4142 < 1, this number is extremely small.

Step 1 — Their sum is an integer
(1 + √2)3000
+
(1 − √2)3000
=
N, an integer

In the two binomial expansions, all terms containing odd powers of √2 cancel. The remaining even powers of √2 are integers. Therefore the sum is an integer.

Step 2 — The conjugate is tiny and positive

Because 3000 is even,

0 < (1 − √2)3000 < 10−100.

One simple bound is:

  1. |1 − √2| < 1/2
  2. (1 − √2)3000 < (1/2)3000 = 2−3000
  3. 23000 > 10100, so 2−3000 < 10−100
Step 3 — So x is just below an integer
x = N − y

with 0 < y < 10−100. Therefore x is less than the integer N by less than one unit in the 100th decimal place.

x = (N − 1).99999999999999999999999999999999999999999999999999...

In fact, the first 100 digits after the decimal point are all 9.

Answer
The 100th digit after the decimal point is
9