Macedonian Mathematical Competitions

Let us assume that $A = 444\ldots488\ldots89$, i.e. $A = \overline{444\ldots488\ldots89}$.

$$\text{Then } A = 4 \cdot \overline{111\ldots1} \cdot 10^n + 8 \cdot \overline{111\ldots1} + 1.$$
$$\text{Now } \overline{111\ldots1} = 10^{n-1} + 10^{n-2} + \dots + 10^2 + 10 + 1 = \frac{10^n - 1}{10 - 1} = \frac{10^n - 1}{9}.$$
$$\text{Then } A = \frac{4}{9}(10^n - 1)10^n + \frac{8}{9}(10^n - 1) + 1 = \frac{4}{9}10^{2n} + \frac{4}{9}10^n + \frac{1}{9} = \left(\frac{2 \cdot 10^n + 1}{3}\right)^2.$$
The sum of the digits of $2 \cdot 10^n + 1$ is equal to $3$, then $2 \cdot 10^n + 1$ is divisible by $3$.