Radical with pattern

Question

Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.

Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.

Without using a calculator, evaluate $\sqrt{a − b}$.

Answer 1

$a= \overbrace{11\ldots11}^{2018} = \dfrac{10^{2018}-1}{9}$

$b= \overbrace{22\ldots22}^{1009} = 2\cdot\dfrac{10^{1009}-1}{9}$

$\Rightarrow a-b = \dfrac{10^{2018}-1}{9}-2\cdot\dfrac{10^{1009}-1}{9} = \dfrac{10^{2018}-2\cdot10^{1009}+1}{9} = \left(\dfrac{10^{1009}-1}{3} \right)^2$

$\Rightarrow \sqrt{a-b} = \dfrac{10^{1009}-1}{3} = \overbrace{33\ldots33}^{1009}$

Answer 2

$$a=\sum_{i=0}^{2017}10^i $$ $$=\frac {10^{2018}-1}{10-1} $$ $$b=2\sum_{i=0}^{1008}10^i $$ $$=2\frac {10^{1009}-1}{10-1}$$

$$9 (a-b)=10^{2018}-2.10^{1009}+1$$ $$=(10^{1009}-1)^2$$

then

$$\sqrt {a-b}=\frac {10^{1009}-1}{3} $$

$=333...333.$ (1009 consecutive times).

Answer 3

Let \begin{eqnarray*} N=\sum_{i=0}^{1008} 10^i =\underbrace{11\cdots 1}_{\text{1009 ones}} \\ M=10^{1009}+1 \end{eqnarray*} note that $M-2=9N$ and
\begin{eqnarray*} a=NM \\ b=2N \end{eqnarray*} So $a-b=9N^2$ and \begin{eqnarray*} \sqrt{a-b} = 3N=\underbrace{\color{red}{33\cdots 3}}_{\text{1009 threes}}. \end{eqnarray*}

Answer 4

\begin{align} a &= \frac{10^{2018}-1}{9} \\ b &= \frac{2(10^{1009}-1)}{9} \\ a-b &= \frac{10^{2018}-2\times10^{1009}+1}{9} \\ &= \frac{(10^{1009}-1)^{2}}{9} \\ \end{align}

Can you proceed?