Find the digit at the $1000$th position at the right of the decimal point of the number $(8+\sqrt{63})^{2012}$

I took this problem from a Mexican Math Olympiad called Galois-Noether. It's the last problem here. The options are $1$, $3$, $7$ and $9$ and the correct one is $9$.

I don't know how to solve this. I know that if we write $(8+\sqrt{63})^{2012}$ as $n.d_1d_2\ldots d_{1000}\ldots$, where $n$ is the integer part of the number and $d_i=0,1,\ldots,9$, then we'll get

$$10^{1000}(8+\sqrt{63})^{2012}=nd_1d_2\ldots d_{1000}.d_{1001}\ldots$$

So all we have to do is to take the integer part of $10^{1000}(8+\sqrt{63})^{2012}$ and then take it $\;\text{mod}\,10$. But, unless you use a computer, this seems too complicated.

One thing I think that'll might help is the fact that $\sqrt{63}$ is ''close'' to $\sqrt{64}=8$. But I can't see how to use this.

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    $\begingroup$ who voted to close this?? $\endgroup$ – symplectomorphic May 15 '16 at 5:11
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    $\begingroup$ $\sqrt63=7.938$ $\endgroup$ – Takahiro Waki May 15 '16 at 7:20
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    $\begingroup$ ax = 6.95960347684E-2419 by calculater.2418th the decimal degit is 9,2419th's is 3. $\endgroup$ – Takahiro Waki May 15 '16 at 7:37

As André Nicolas has kindly pointed out, $$(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$$ is equal to an integral value. Also, $(8-\sqrt{63})^{2012}$ is equal to a fractional value. Thus, we can write: $$I + f = (8+\sqrt{63})^{2012}$$ $$f` = (8-\sqrt{63})^{2012}$$ Where I means integer while f means fractional part. Addding the two, we get: $$I + f + f` = (8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$$ Now, as both f and $f`$ are between 0 and 1, their sum will be between 0 and 2. However, we know that $I + f + f`$ is equal to a integral value and thus $f + f`$ can only equal to 1. Hence: $$f + f` = 1$$ $$f = 1 - f`$$ As the value of $f`$ is quite small, $1-f`$ will give a number very close to 1, i.e $0.999.....$

  • $\begingroup$ Thanks man, it was very clarifying! $\endgroup$ – Larara May 15 '16 at 5:04
  • $\begingroup$ This is absurdly overcomplicated and certainly not the best answer to this question. There is no point to writing $I+f$. The real point is that the desired number is an integer minus $f'$, as you wrote it, but $f'$ is extremely small. $\endgroup$ – symplectomorphic May 15 '16 at 5:06
  • $\begingroup$ @symplectomorphic Agreed. But I believe this makes it easier to understand. Also, when finding out that the value is equal to an integer, you're going to have to use binomial which makes it long enough. Once you're clear with the concept, you can quickly make assumptions and do majority of the steps in your mind. $\endgroup$ – Gummy bears May 15 '16 at 5:08
  • $\begingroup$ @Gummy bears: I disagree. It obscures the key insight, which has nothing to do with your $I$ and $f$. Roman83's answer better distills André's hint. $\endgroup$ – symplectomorphic May 15 '16 at 5:10
  • $\begingroup$ @symplectomorphic I did get carried away xD This method is better when we're supposed to find whether the integral part of the number is odd or even. $\endgroup$ – Gummy bears May 15 '16 at 5:11


(i) $(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}$ is an integer. One way to prove this is to use the binomial theorem.

(ii) $(8-\sqrt{63})^{2012}$ is positive and quite a bit less than $10^{-1000}$. (The fact that $\sqrt{63}$ is close to $8$ is useful for this part).

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    $\begingroup$ @Larara Check my answer. I think it should clarify it. $\endgroup$ – Gummy bears May 15 '16 at 5:01
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    $\begingroup$ @Larara: So our number is a very tiny bit less than a positive integer. It's thousandth digit after the decimal point is therefore $9$. By the way, that power is quite a bit less than $10^{1012}$, indeed quite a bit less than $10^{-2000}$. If we were asked for the $2000$-th digit after the decimal point the answer would still be $9$. $\endgroup$ – André Nicolas May 15 '16 at 5:11
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    $\begingroup$ You are welcome. To get a quickie estimate of $(8-\sqrt{63})^{2012}$, note this is $(8+\sqrt{63})^{-2012}$, about $16^{-2012}$, which is less than $2^{-8000}$. But $2^{10}\approx 10^3$, so we are around $10^{-2400}$. $\endgroup$ – André Nicolas May 15 '16 at 5:20
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    $\begingroup$ 2418 was computed by calculater. $\endgroup$ – Takahiro Waki May 15 '16 at 7:33
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    $\begingroup$ @Larara $(x - \sqrt{y})^n ~ 10^k < 1$ is equivalent to $x - \sqrt{y} < 10^{-k/n}$, which is the case here as $8 - \sqrt{63} \approx 0.06$ and $10^{-1000/2012} \approx 0.3$ $\endgroup$ – DanielV May 15 '16 at 9:42

1) $(8-\sqrt{63})^{2012}=N.0000000...00ABC...$

2) $(8+\sqrt{63})^{2012}+(8-\sqrt{63})^{2012}=A \in Z$


3) $(8+\sqrt{63})^{2012}=K.9999999...99DCE...$


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