Prove that for any $n\in\mathbb{N}$ there exists a number $m\in\mathbb{N}$ such that the decimal representation of $m^2$ has $n$ ones at the beginning and some combination of $n$ ones and twos at the end.
This is the high school olympiad problem and nothing else is given. I can only show some results on the first part of the problem. Here is what I have so far:
Given $n$, there exists an integer $p$ such that the decimal representation of $p$ starts with $n$ ones.
Proof. Take $c = \underbrace{111\!\dots\!11}_{n \text{ times}}\cdot 10^k$ and $d = \underbrace{\!111\!\dots\!1\!\!}_{n-1 \text{ times}}\ 2\!\cdot 10^k = c + 10^k$, $k$ is a positive integer. Then $$ \sqrt{c} - \sqrt{d} = \dfrac{c-d}{\sqrt{c} + \sqrt{d}} \approx \frac{10^{k}}{2\cdot 10^{(n-1+k)/2}}> \frac{10^k}{2\cdot 10^{k-1}}>1 $$ The last estimate holds as long as $k>n+1$, because $10^{(n-1+k)/2}<10^{k-1} \implies \frac{n+1}{2}<\frac{k}{2}$.
In this way, there is an integer number $p$ such that $\sqrt{c}<p<\sqrt{d}$ and $p^2 = \underbrace{111\!\dots\!11}_{n \text{ times}}\cdot 10^k + S$, where $0<S<10^k$.
How to approach the second half, the one about $n$-long combination of 1 and 2 at the end of $m^2$?