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?

  • 1
    $\begingroup$ Do you know what the first digits (the ones before the decimal point and the 1004 first digits after it) look like? $\endgroup$
    – Fabian
    Aug 19, 2014 at 7:37

2 Answers 2


Note that $N=\frac{10^{2008}-1}{9}$. One can use the Taylor series of $\sqrt{x}$ at $a=10^{2008}$ to solve this problem. More precisely, let $f(a)=\sqrt a$,

$$f(x)=f(a)+(x-a)f'(a)+\frac{f''(a)}2(x-a)^2+\cdots$$ Or $$\sqrt{9N}=10^{1004}-\frac{1}{2\cdot 10^{1004}}-\frac{3}{4\cdot 10^{3\cdot 1004}}+\cdots$$ That means $$3\sqrt{N}=10^{1004}-5\cdot 10^{-1005}-o(10^{-3000}).$$ The answer is obvious from here.


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.


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