When written in decimal notation, every square number has at most 1000 digits that are not 0 or 1. True or false? 
When written in decimal notation, every square number has at most $1000$ digits that
  are not $0$ or $1$. True or false? 

This question is from an admissions quiz, so no calculators should be used.
 A: This is so grossly false that I don't think it is worth constructing a very detailed counterexample. The ratio of successive squares $\frac{(n+1)^2}{n^2}$ decreases and tends to$~1$, so for any $\epsilon>0$ there is an (easily found) $~N\in\Bbb N$ such that $1<\frac{(n+1)^2}{n^2}<1+\epsilon$ for all $n\geq N$. Now take a number $m$ given by any sequence of digits that violates your condition ($1001$ digits $7$ will do) and find $N$ corresponding to $\epsilon=\frac1{10m}$. Now take $k\in\Bbb N$ sufficiently large so that $10^km>N^2$ and put $n=\lfloor\sqrt{10^km}\rfloor$ (one has $n\geq N$). Then one has $10^km<(n+1)^2<10^k(m+\frac1{10})$ so the number $(n+1)^2$ starts with the digits of $m$.
The same argument shows that any finite sequence of digits occurs (infinitely often) as the initial digit sequence of a square. It can also be adapted easily to many other kinds of numbers with a similar density.
A: $$(10^n+n)^2=10^{2n}+2n \cdot 10^n+n^2$$
So each even number can appear as pattern in a square number. If one chooses $n=5m$ then $2n=10m$ and so each $m \in \mathbb{N}$ appears as digit pattern in the above sequence.
A: Here's the first thing I dug out of a cobweb covered corner in the old brain. Similar in spirit to that of Oliver's
$$
\begin{aligned}
35^2&=1225\\
335^2&=112225\\
3335^2&=11122225\\
\cdots&=\cdots
\end{aligned}
$$
Here $x_n=333\ldots5=(10^n+5)/3$. Therefore
$$
x_n^2=\frac{10^{2n}+10^{n+1}+25}9.
$$
The first $(n-1)$ digits of that are equal to those of $10^{2n}/9$, i.e. all ones. Similarly the next $n$ digits are all twos.

For $111\ldots11^2$ one can say something precise also. This problem shows that if the number of ones in the string of digits is $n=9q+r, 0\le r<9,$ then the some of the digits of the square is $81q+r^2$. As the number of digits in the square is about $2n\approx 18q$ the average digit has to be about $81/18=9/2$, and thus the number of non-$0/1$s grows without a bound.
A: This is how I did it, but I'm sure there are better ways.
Let $x = 10^n - 1$ with $n\in\mathbb{Z}$, then $x^2 = 10^{2n} - 2\times10^n + 1 = 999\cdots9998000\cdots001 $
Let $n>1001$  so we have more than $1000$ nines, then we've found a square number with more than $1000$ digits that are not $0$ or $1$, so it's false.
