63rd Czech and Slovak Mathematical Olympiad

The number under consideration can be expressed as follows: $$\begin{array}{rl}& (10^{2n-1}+10^{2n-2}+\cdots +10^{n+1})+2\cdot 10^n+8\cdot (10^{n-1}+10^{n-2}+\cdots +10^2)+96\\ & =10^{n+1}\cdot \frac{10^{n-1}-1}{9}+2\cdot 10^n+8\cdot 10^2\cdot \frac{10^{n-2}-1}{9}+96\\ & =\frac{10^{2n}-10^{n+1}+18\cdot 10^n+800\cdot 10^{n-2}-800+9\cdot 96}{9}\\ & =\frac{10^{2n}+16\cdot 10^n+64}{9}={\left(\frac{10^n+8}{3}\right)}^2.\end{array}$$ As required, we have obtained a perfect square, because the number $10^n+8$ is divisible by 3, as the sum of its digits equals 9.

$1296=36^2,112896=336^2,11128896=3336^2,\dots$ we easily guess that for each $n\ge 2$, $$\underset{n-1}{\underbrace{1\dots 128}}\underset{n-2}{\underbrace{896}}=\underset{n-1}{\underbrace{33\dots 36^2}}.$$ The exact proof can be done by using the usual multiplication scheme: $$\begin{array}{r}333\dots 3336\\ \times 333\dots 3336\\ 2000\dots 0016\\ 10000\dots 008\\ 100000\dots 08\\ 1000000\dots 8\\ ⋮\\ 1\dots 000008\\ 10\dots 00008\\ 100\dots 0008\\ 111\dots 112888\dots 8896\end{array}$$

Both (identical) factors are $n$-digit, hence an $(n+1)$-digit number stands in each of the $n$ rows between the two delimiting lines. From this fact, it is easy to determine the values of digits (including the numbers of appearances) in the resulting product.