<s>[INST] 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 $\sqrt{a − b}$.
 [/INST] $a= \overbrace{11\ldots11}^{2018} = \dfrac{10^{2018}-1}{9}$
$b= \overbrace{22\ldots22}^{1009} = 2\cdot\dfrac{10^{1009}-1}{9}$
$\Rightarrow a-b = \dfrac{10^{2018}-1}{9}-2\cdot\dfrac{10^{1009}-1}{9} =  \dfrac{10^{2018}-2\cdot10^{1009}+1}{9} = \left(\dfrac{10^{1009}-1}{3} \right)^2$
$\Rightarrow \sqrt{a-b} = \dfrac{10^{1009}-1}{3} = \overbrace{33\ldots33}^{1009}$
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