Egorov A.A.

Calculate the square root of the number 0.111...111 (100 ones) with an accuracy of a) 100; b) 101; c)* 200 decimal places.
Let's provide a solution that allows calculating this root with an accuracy of 301 digits after the decimal point.

Transform the given number as follows:

$$
\frac{1}{3} 100=\frac{1}{3} \sqrt{1-10^{-100}}
$$

Then

$(1)$

$$
\sqrt{1-10^{-100}} \approx 1 - 0.5 \cdot 10^{-100} - b \cdot 10^{-200}
$$

which is equivalent to

(1) $\quad 0 > (0.25 - 2b) + b \cdot 10^{-100} + b^2 \cdot 10^{-200}$.

To satisfy inequality (eq:254.4), it is sufficient to set $b = 0.125 + 10^{-101}$.

We obtain the following estimate:

(1)

$\sqrt{1-10^{-100}} > 1 - 0.5 \cdot 10^{-100} - 0.125 \cdot 10^{-200} - 10^{-301}$

From estimates (eq:254.3) and (eq:254.5), it follows that

$$
0.10049987498 < \sqrt{1-10^{-100}} < 0.10049987499
$$

From this, we get

$$
100 = 0.100100 ; 98 \ldots
$$

where all 301 digits after the decimal point are correct.

A similar solution to this problem was submitted by our reader S. Domoryad.