# Ask Uncle Colin: A 100th digit of a 500th power

Dear Uncle Colin,

How on earth would I work out the 100th digit after the decimal point in $(1 + \sqrt{2})^{500}$? The would require an enormous calculation!

- Perhaps A Simplification Could Answer Logically

Hi, PASCAL, and thanks for your message! The answer to this lies in one of my favourite tricks.

If you work out the binomial expansions of $(1+x)^{2n}$ and of $(1-x)^{2n}$, you find that the terms in either expression with even powers of $x$ are equal, and those with odd powers of $x$ differ only in their sign. Adding the two together gives an expression that has only even powers of $x$ in it.

In particular, if $x = \sqrt{2}$, $(1+x)^{2n} + (1-x)^{2n}$ is an integer, because even powers of $\sqrt{2}$ are powers of 2.

The second trick is to note that $|1-x| < 1$ here - so $(1-x)^{2n}$ is very small.

### Whoosh

“Of course, sensei, this has your fingerprints all over it.”

“I do *not* leave fingerprints.”

“Put the pineapple juice down and get on with it.”

“$\sqrt{2} - 1$ is less than a half, so $(1-\sqrt{2})^{500}$ is smaller than $2^{-500}$.”

“Unusually obvious so far.”

“Since $2^{-10} < 10^{-3}$, $2^{-500} < 10^{-150}$.”

“Is that the best you’ve got?”

“Fine. It’s about five-sixths of a half. $\left(1 - \frac{1}{6}\right)^{11}$ is about $e^{-2}$, so to the power of 500 is $e^{-91}$ish. And $e^{-91}$ is about $10^{-39}$, so I’ll say it’s about $10^{-190}$.”

“My sources say $4 \times 10^{-192}$.”

“Your *sources* had better watch out I don’t spill pineapple juice all over their keyboards.”

### So where were we?

To recap, two major points: $(1+\sqrt{2})^{500} + (1-\sqrt{2})^{500}$ is an integer (let’s call it $N$), and $(1-\sqrt{2})^{500}$ is tiny – let’s call it $\epsilon$ – and $\epsilon$ has well over a hundred zeros after the decimal point before you hit the first non-zero decimal digit.

That means $(1 + \sqrt{2})^{500}$ is $N - \epsilon$, and after its decimal point, it has well over a hundred 9s. **The 100th digit after the dot is therefore 9.**

### Where else does this trick work?

Similar tricks will work for any $(\sqrt{a}+ \sqrt{b})^{2n}$ as long as $n$ is very big and $|\sqrt{a} - \sqrt{b}| < 1$. This turns out to be equivalent to the difference between the arithmetic mean and the geometric mean of $a$ and $b$ being less than $\frac{1}{2}$.

Hope that helps!

- Uncle Colin