The 100th Decimal Digit of (1 + sqrt 2)^3000

Worked Solution

How to Think About It: A 3000th power of an irrational number looks hopeless until you notice its algebraic conjugate. Adding $(1 - \sqrt{2})^{3000}$ kills every $\sqrt{2}$ term and leaves an integer; and because $|1 - \sqrt{2}| < 1$, the conjugate term is astronomically small. That pins down the fractional part almost exactly.

Quick Estimate: $\log_{10} |1 - \sqrt{2}| = \log_{10}(0.4142) \approx -0.383$, so $(1 - \sqrt{2})^{3000} \approx 10^{-1148}$. The number is an integer minus $10^{-1148}$, so its first 1148 decimal digits are all 9. In particular the 100th one is 9.

Formal Solution:

*Step 1 -- The conjugate sum is an integer.* By the binomial theorem,

$$(1 + \sqrt{2})^n + (1 - \sqrt{2})^n = \sum_{k=0}^{n} \binom{n}{k} \left(\sqrt{2}\right)^k \left(1 + (-1)^k\right) = 2 \sum_{k \text{ even}} \binom{n}{k} 2^{k/2},$$

which is an integer. Call it $N$ for $n = 3000$.

*Step 2 -- The conjugate term is tiny and positive.* Since $1 - \sqrt{2} \approx -0.4142$ and $3000$ is even,

$$0 < (1 - \sqrt{2})^{3000} = (\sqrt{2} - 1)^{3000} < (0.5)^{3000} = 2^{-3000} < 10^{-900}.$$

(In fact it is about $10^{-1148}$.)

*Step 3 -- Read off the digits.* Therefore

$$(1 + \sqrt{2})^{3000} = N - \varepsilon, \qquad 0 < \varepsilon < 10^{-900}.$$

A number of the form $N - \varepsilon$ with $N$ an integer and $0 < \varepsilon < 10^{-900}$ has decimal expansion $(N - 1).999\ldots9\ldots$ with at least 900 consecutive 9s after the decimal point. The 100th digit after the decimal point is therefore 9.

*Remark.* If the exponent were odd, $(1 - \sqrt{2})^n$ would be negative, the number would be $N + \varepsilon$, and the answer would be 0 instead. The parity of the exponent matters.

Answer: The 100th digit after the decimal point is $9$.