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

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    Are you sure you didn't mean $(1+\sqrt{2})^{2000}$ or something like that? – 2017-01-10
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    I'm assuming you have already tried Wolfram-Alpha-ing it up? – 2017-01-10
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    sorry guys, just edited it – 2017-01-10
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    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. – 2017-01-10

2 Answers 2

7

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?

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    I don't really understand, why did you add $(1 - \sqrt 2)^{3000}$ , this part is essentially 0, correct ? – 2017-01-10
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    ^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? – 2017-01-10
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    it's gonna be all 9s ? thanks. – 2017-01-10
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    if $( 1 - \sqrt 2 )^{3000}$ is a positive number, why do you say "minus" ? – 2017-01-10
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    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. – 2017-01-10
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    that makes sense, thanks. – 2017-01-10
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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

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    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?) – 2017-01-10
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    ... 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. – 2017-01-10
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    $(1 + \sqrt{2})^k = \sum {3000 \choose k} \sqrt{2}^k$. $(1 - \sqrt{2})^k = \sum (-1)^k {3000 \choose k} \sqrt{2}^k$. 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. – 2017-01-10