TITLE: Find the 1005th digit after the decimal point expansion of the square root of N.
QUESTION [5 upvotes]: Let $N$ be the positive integer with $2008$ decimal digits, all of them $1$. That is, $N=1111...1111$, with $2008$ occurrences of the digit $1$. Find the $1005th$ digit after the decimal point expansion of $\sqrt{N}$.
The proof given simply states two values and shows that their squares are greater than and less than $N$, and uses this to show that they are less than and greater than $\sqrt{N}$ and so the $1005th$ digit is $1$. This was from a calculator free exam so I don't see how this could have possibly been done, does anyone have any ideas?

REPLY [2 votes]: Maybe if we write $N=\lfloor 10^{2008}\cdot 0.\overline{111}\rfloor=\dfrac{10^{2008}-1}{9}$. Then we must have
$$
\sqrt{N}\approx\sqrt{N+\tfrac19}=\sqrt{\frac{10^{2008}}{9}}=\frac{10^{1004}}3=10^{1004}\cdot 0.\overline{333}
$$
and with the derivative of the square root function $f(x)=\sqrt x$ being $f'(x)=\frac{1}{2\sqrt x}$ the difference between $\sqrt{N}$ and $\sqrt{N+\frac19}$ will be approximately
$$
\frac19\cdot f'(N)=\frac19\cdot\frac{1}{2\sqrt N}\approx \frac19\cdot\frac{3}{2\cdot 10^{1004}}=\frac{1}{6\cdot 10^{1004}}=1.\overline{666}\cdot 10^{-1005}
$$
So $\sqrt N$ must be extremely close ($f(x)$ is almost linear for so large values of $x$) to
$$
\sqrt N\approx 10^{1004}\cdot 0.\overline{333}-1.\overline{666}\cdot 10^{-1005}
$$
Note that $3.\overline{333}-1.\overline{666}=\frac{10}{3}-\frac{10}{6}=\frac{10}{6}=1.\overline{666}$ so the 1005th digit must be $1$. However, Quang Hoang applied a more precise method for controlling the size of the error. Also Taylor Expansion is a more general tool. I will leave this answer here, nonetheless, since it is always entertaining to have different approaches represented.