Q:

Prove that integer $n$ exists such that $n^2$ begins with $201120122013$.

I've found a few different formulations of the problem where the given digits are different, so my guess is that it actually works for any array of integers. But I don't know how to solve it, nor where to start. I'm not that good in number theory.

A:

Consider $y = 10^k\sqrt{201120122013}$ then $\lceil y\rceil^2 < (y+1)^2 = 10^{2k} 201120122013 + 2y + 1$.  Now $2y+1<10^{k+12}$ so if $k>12$ the leading digits of $\lceil y\rceil^2$ are 201120122013.

A:

$$ 44846418141586293^2 = 2011201220130000177943626365481849 $$

A:

You're right that there's a general procedure for this - it's based on approximations of the square root.  Suppose we have a number $n^2$ of length $d$ digits - in other words, $10^{d-1}\leq n^2 \leq 10^d$.  Then saying that the first $12$ digits of $n^2$ are $201120122013$ is the same as saying that the first $12$ digits (after the decimal point) of $n^2/10^d$ are $0.201120122013$; or in other words, that $0.201120122013 \leq n^2/10^d \le 0.201120122013+10^{-12}$.
Now, let $t = \sqrt{0.201120122013} = .44846418141586\ldots$ and consider the numbers $t_i = \lceil10^i\cdot t\rceil$ - these correspond to taking longer and longer 'overestimates' of the digits of $t$; for instance, $t_1 = 5, t_2 = 45, t_3 = 449,\ldots$  Then we know that $0\leq t_i-10^i\cdot t\lt 1$ (by the definition of the ceiling function), so we know that $0\leq t_i^2-10^{2i}\cdot t^2 = (t_i-10^i\cdot t)\cdot (t_i+10^i\cdot t) \lt t_i+10^i\cdot t \lt 2(10^i\cdot t+1)$; since $t$ is less than $1$ then the last value is certainly less than $2\cdot 10^i$.  But we can divide this by $10^{2i}$ to get $t^2\leq t_i^2/10^{2i}\lt t^2+ 2\cdot10^{-i}$ - or in other words, $0.201120122013 \leq t_i^2/10^{2i} \le 0.201120122013+2\cdot10^{-i}$ - and all we have to do to get this to match up with our original inequality is to take an $i$ such that $2\cdot 10^{-i}$ is even less than $10^{-12}$ - for instance, $i=13$ will do.  This gives us an answer that $t_{13} = 4484641814159$ squares to $t_{13}^2=20112012201303326692877281$.

