Radical with pattern 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 $\sqrt{a − b}$.
 A: \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?
A: $$a=\sum_{i=0}^{2017}10^i $$
$$=\frac {10^{2018}-1}{10-1} $$
$$b=2\sum_{i=0}^{1008}10^i $$
$$=2\frac {10^{1009}-1}{10-1}$$
$$9 (a-b)=10^{2018}-2.10^{1009}+1$$
$$=(10^{1009}-1)^2$$
then
$$\sqrt {a-b}=\frac {10^{1009}-1}{3} $$
$=333...333.$ (1009  consecutive times).
A: 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*}
A: $a= \overbrace{11\ldots11}^{2018} = \dfrac{10^{2018}-1}{9}$
$b= \overbrace{22\ldots22}^{1009} = 2\cdot\dfrac{10^{1009}-1}{9}$
$\Rightarrow a-b = \dfrac{10^{2018}-1}{9}-2\cdot\dfrac{10^{1009}-1}{9} =  \dfrac{10^{2018}-2\cdot10^{1009}+1}{9} = \left(\dfrac{10^{1009}-1}{3} \right)^2$
$\Rightarrow \sqrt{a-b} = \dfrac{10^{1009}-1}{3} = \overbrace{33\ldots33}^{1009}$
