1. Zadatak

Označimo sa $x$ traženi broj, i neka je $a=\underbrace{11\ldots1}_{n}$. Tada je

$$10^n=\underbrace{99\ldots9}_{n}+1=9a+1$$

pa dobivamo

$$\underbrace{44\ldots4}_{2n}=4a\cdot10^n+4a=4a(9a+2),$$ $$\underbrace{11\ldots1}_{n+1}=10a+1,$$ $$\underbrace{66\ldots6}_{n}=6a.$$

Zato je

$$\begin{aligned}x^2&=4a(9a+2)+10a+1-6a\\&=36a^2+12a+1=(6a+1)^2\end{aligned}$$

Odavde slijedi $x=6a+1=6\cdot\underbrace{11\ldots1}_{n}+1=\underbrace{66\ldots6}_{n-1}7$.