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As I was cleaning up my desk, I found my Calculus exam from almost a year ago. I remember there was only a bonus task that required either a tad more wit, either a bit more time. It goes like this :


$$ \text{Find the 1000-th decimal of }\underbrace{\sqrt{1111...111}}_{1998 \text{ times}} . $$


I remember noticing $11 = \frac{10^2-1}{9}$, building up a general case upon this observation and representing it via series using the binomial theorem, but nothing actually led me to the actual answer.

Any ideas ?

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  • $\begingroup$ You could compute this mess, using Maple. However it would be a cumbersome task. $\endgroup$ – Aleksandar Oct 4 '15 at 20:27
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    $\begingroup$ Note this was a single task out of a total of $10$, in an exam that allowed no other instruments than pen and paper. I'm interested in a solution, I couldn't care less about the actual digit $\endgroup$ – Victor Oct 4 '15 at 20:29
  • $\begingroup$ Only as a bonus task. Other subjects were quite approachable by using conventional techniques taught in Calculus 3. $\endgroup$ – Victor Oct 4 '15 at 20:37
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    $\begingroup$ Here's some food for thought... Compare $\sqrt{11}\approx 3.3$ with $\sqrt{1111}\approx33.33$ with $\sqrt{111111}\approx333.333$. Then we expect that $3$ is our answer. $\endgroup$ – zahbaz Oct 4 '15 at 20:58
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    $\begingroup$ Using an unproved rule that I have observed, I would say that the result is (0.33333333...+ the number 3 repated 333 times (as in 33333.....to 333 digits). So the last integer digit would be 3. $\endgroup$ – NoChance Oct 4 '15 at 20:58
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So you get $$ \frac13\sqrt{10^{1998}-1}=\frac13·10^{999}·\sqrt{1-10^{-1998}} $$ and by the binomial series $$ \sqrt{1+x}=1+\frac12x -\frac18x^2+\frac1{16}x^3-\frac{5}{128}x^4\pm… $$ it is only the first three terms that influence the first 4000 or so digits leading to \begin{align} \frac13·10^{999}·(1-5·10^{-1999}-1.25·10^{-3997}-…) &=\frac13·10^{999}·0.\underbrace{99…9}_{1998}4\underbrace{99…9}_{1997}874999… \\ &=10^{999}·0.\underbrace{33…3}_{1998}1\underbrace{66…6}_{1997}624999 \end{align} I'll leave you to find out where the 1000 decimal digit after the point comes to lie. To compare: \begin{align} \sqrt{999999}&=999.99949999987499993749996093747265622949217138670565794807… \\ \sqrt{111111}&=333.33316666662499997916665364582421874316405712890188598269… \\ \sqrt{99999999}&=9999.99994999999987499999937499999609374997265624979492187338… \\ \sqrt{11111111}&=3333.33331666666662499999979166666536458332421874993164062446… \end{align}

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