# Find the 100th digit of $( 1 + \sqrt 2 )^{3000}$

What is the 100th digit to the right of the decimal point in the decimal representation of $$( 1 + \sqrt 2 )^{3000}\ ?$$

• Are you sure you didn't mean $(1+\sqrt{2})^{2000}$ or something like that? – JimmyK4542 Jan 10 '17 at 4:30
• I'm assuming you have already tried Wolfram-Alpha-ing it up? – Frank Jan 10 '17 at 4:31
• sorry guys, just edited it – szd116 Jan 10 '17 at 4:34
• The way the question is asked hints that this number must be very very close to a simple rational, otherwise computing the $100^{th}$ digit would be a nightmare. – Yves Daoust Jan 10 '17 at 9:09
• It is also useful to know that (somewhat by guessing, but I mean $\vert 1-\sqrt 2\vert$ is $<0.5$ , it's impossible $0.5^{10}>0.1$) $$(1-\sqrt 2)^{10}<0.1^{1}$$ then it also makes sense $$(1-\sqrt2)^{1,000}<0.1^{100}$$ We can infer that $(1-\sqrt2)^{3,000}<<0.1^{100}$ so it shall not affect the 100th digit to the right of decimal "directly", like 1-0.000...01 – Ruizhen Mai Aug 8 '19 at 14:19

Hints: $(1+\sqrt{2})^{3000}+(1-\sqrt{2})^{3000}$ is an integer and $(1-\sqrt{2})^{3000}$ is a small positive number.

Can you prove these two facts and use them to get the answer?

• I don't really understand, why did you add $(1 - \sqrt 2)^{3000}$ , this part is essentially 0, correct ? – szd116 Jan 10 '17 at 4:37
• ^Yes it is really small. So $(1+\sqrt{2})^{3000}$ is an integer minus a very small number. What does that tell you about the first several digits to the right of the decimal point? – JimmyK4542 Jan 10 '17 at 4:42
• it's gonna be all 9s ? thanks. – szd116 Jan 10 '17 at 4:44
• if $( 1 - \sqrt 2 )^{3000}$ is a positive number, why do you say "minus" ? – szd116 Jan 10 '17 at 4:50
• We know that $N := (1+\sqrt{2})^{3000}+(1-\sqrt{2})^{3000}$ is a positive integer. You can prove this by using the binomial theorem, and noticing that all the terms with an odd power of $\sqrt{2}$ cancel out. So, $(1+\sqrt{2})^{3000} = N-(1-\sqrt{2})^{3000}$ is a positive integer minus a very small positive number. – JimmyK4542 Jan 10 '17 at 4:52

The way I understand it now is that since
$(1+\sqrt{2})^{2} + (1 - \sqrt{2})^{2} = 6$
$\therefore$ $(1+\sqrt{2})^{3000} + (1 - \sqrt{2})^{3000} = N$ (N is an integer)
since $(1 - \sqrt{2})^{3000}$ would be a very small number so $(1+\sqrt{2})^{3000} = N - 0.0000...001$ so the 100th digit to the right of the decimal point should be 9

• Mmmm.... I don't like you saying "a small number is 0.00000........0001". $(1-\sqrt{2})^{3000}$ is quite likely irrational and will not have a final digit of 1. In fact it is easy (but a waste of time) to prove $(1-\sqrt{2})^{3000} \ne (\frac 1{10})^k$ for any integer $k$. Also $(1-\sqrt{2})^{3000}$ is a "very small number" but how small? How do you know its 100th digit is 0. Is it that small? How do you know? (Hint: is $(1- \sqrt{2})^{3000} < (\frac 1{10})^{100}$? How do you know?) – fleablood Jan 10 '17 at 18:31
• ... also $a + b = 6$ does not mean $a^k + b^k$ is an integer. $2.5 + 3.5 = 6$ and $2.5^{3} + 3.5^{3} = 58.5$ which is not an integer. ... Sorry, I'm being a hard-ass but... you are on the right track. But you can't afford sloppy corner cutting. – fleablood Jan 10 '17 at 18:37
• $(1 + \sqrt{2})^k = \sum {3000 \choose k} \sqrt{2}^k$. $(1 - \sqrt{2})^k = \sum (-1)^k {3000 \choose k} \sqrt{2}^k$. So $(1 + \sqrt{2})^k+(1 - \sqrt{2})^k = \sum {3000 \choose k}[ \sqrt{2}^k + (-1)^k\sqrt{2}^k]$. $(-1)^k = -1$ if $k$ is odd and $=1$ if $k$ is even so $(1 + \sqrt{2})^k+(1 - \sqrt{2})^k =$$\sum_{k \text{ is even}} {3000 \choose k}*2*\sqrt{2}^k$$= \sum_{k \text{ is even}} {3000 \choose k}*2*2^{k/2}$ which is an integer. That's the real reason. – fleablood Jan 10 '17 at 18:42