4. Determine all 200-digit natural numbers (in decimal notation) that are squares of natural numbers and start with 99 nines.
4. Numbers that have 200 digits and start with 99 nines are numbers between $10^{200}-10^{101}$ and $10^{200}-1$.

Since $\left(10^{100}\right)^{2}=10^{200}$, it follows that the desired numbers are $10^{100}-1$ and smaller.

We claim that these numbers are $10^{100}-1, 10^{100}-2, 10^{100}-3, 10^{100}-4$, and $10^{100}-5$.

It is sufficient to check that $\left(10^{100}-5\right)^{2} \geq 10^{200}-10^{101}$ and $\left(10^{100}-6\right)^{2} < 10^{200}-10^{101}$:
\[
\begin{gathered}
\left(10^{100}-5\right)^{2} = \\
= 10^{200} - 10 \cdot 10^{100} + 25 \\
= 10^{200} - 10^{101} + 15 \geq 10^{200} - 10^{101} \\
\left(10^{100}-6\right)^{2} = \\
= 10^{200} - 12 \cdot 10^{100} + 36 \\
= 10^{200} - 10^{101} - 2 \cdot 10^{100} + 36 < 10^{200} - 10^{101}
\end{gathered}
\]

MINISTRY OF EDUCATION AND SPORT OF THE REPUBLIC OF CROATIA

CROATIAN MATHEMATICAL SOCIETY

# MATHEMATICS