Simplifying $\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$ How do I simplify:
$$\sqrt{\underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}}$$
Should I use modulos or should I factor them?  Or any I suppose to use combinatorics?  Any one have a clue?
 A: Nice question there!
Let $x=\underbrace{11\cdots1}_{n\ 1's}$, then
\begin{align}
\therefore\underbrace{11\cdots1}_{2n\ 1's}-\underbrace{22\cdots2}_{n\ 2's}&=\underbrace{11\cdots1}_{n\ 1's}\times10^n+\underbrace{11\cdots1}_{n\ 1's}-2\times\underbrace{11\cdots1}_{n\ 1's}
\\ 
&=x\times10^n+x-2x
\\ \\
&=x\times10^n-x
\\ \\
&=x(10^n-1)
\\ \\
&=x\times\underbrace{99\cdots9}_{n\ 9's}
\\
&=x\times(9\times\underbrace{11\cdots1}_{n\ 1's})
\\
&=9x^2
\end{align}
\begin{align}
\therefore\text{The original equation is equal to: }\sqrt{9x^2}=3x=\boxed{\underbrace{33\cdots3}_{n\ 3's}}
\end{align}
A: Observe that for integer $m\ge 0,$
   $$\underbrace{11\dots1}_{m\ 1's}=\frac{\underbrace{99\dots99}_{m\ 9's}}9=\frac{10^m-1}9 $$
$$\text{So, }\underbrace{11\dots11}_{2n\ 1's}=\frac{10^{2n}-1}9\text{ and }\underbrace{22\dots22}_{n\ 2's}=2\cdot \underbrace{11\dots11}_{n\ 1's}=2 \cdot\frac{10^n-1}9$$
$$\implies \underbrace{11\dots1}_{2n\ 1's}-\underbrace{22\dots2}_{n\ 2's}=\frac{10^{2n}-1-2(10^n-1)}9=\frac{10^{2n}-2\cdot10^n+1}9=\left(\frac{10^n-1}3\right)^2$$
$$\text{Now as }\underbrace{11\dots1}_{m\ 1's}=\frac{10^m-1}9,$$
$$ \frac{10^n-1}3=3\cdot \frac{10^n-1}9=3\cdot(\underbrace{11\dots11}_{n\ 1's})=\underbrace{33\dots33}_{n\ 3's}$$
