Provide integer values for $a$ and $b$ such that $2003 < a + b \sqrt{2} < 2003.01$.
I. solution. In fact, it doesn't take long to find suitable values of $a$ and $b$ even with a calculator, but we can save time if we examine the number $\sqrt{2}$ to a few decimal places. The task is clearly to find an integer $b$ such that the fractional part of $b \sqrt{2}$ is less than $0.01$.

Approximately: $\sqrt{2} \approx 1.4142136$, the digit in the tenths and thousandths place is 4. This gives the idea to examine the difference $100 \sqrt{2} - \sqrt{2}$, in which we hope that at least two zeros follow the decimal point in the difference.

Indeed: $100 \sqrt{2} - \sqrt{2} \approx 140.0071427$, so for $b=99$, $0<\{b \sqrt{2}\}<0.01$ holds. We only need to find the appropriate $a$ number, which we can easily find knowing $b$: $a=1863$.

The numbers $a=1863$ and $b=99$ satisfy:

$$
2003<1863+99 \sqrt{2}(\approx 2003.007)<2003.01
$$

Kristóf Bérczi (Szeged, SZTE Ságvári Endre Gyakorló Gimnázium, 12th grade)

II. solution. It can be verified that $0<\sqrt{2}-1<\frac{1}{2}$, from which

$$
\begin{aligned}
0<3-2 \sqrt{2} & =(\sqrt{2}-1)^{2}<\frac{1}{4} \\
0<17-12 \sqrt{2} & =(3-2 \sqrt{2})^{2}<\frac{1}{16}<\frac{1}{10} \\
0<577-408 \sqrt{2} & =(17-12 \sqrt{2})^{2}<\frac{1}{100}
\end{aligned}
$$

from which $2003<2580-408 \sqrt{2}<2003.01$. So, for example, $a=2580, b=-408$ is suitable.