\section*{Exercise 1 - 041211}

From a four-digit table, we obtain the following approximate values:

\[
\sqrt[3]{636000} \approx 86.00 \text { and } \sqrt[3]{389000} \approx 73.00
\]

Therefore, \(z=\sqrt[3]{636000}-\sqrt[3]{389000} \approx 13\).

Without using any further table, it should be decided whether \(z\) is greater, smaller, or equal to 13.
}

We consider the three possibilities \(z>13\), \(z=13\), and \(z<13\). Since \((-13-\sqrt{25277}) / 20 < 0\), the second factor in the last line is positive, so the relation is equivalent to

\[
\begin{gathered}
0 \gtreqless a-\frac{-13+\sqrt{25277}}{20} \Leftrightarrow -13+\sqrt{25277} \gtreqless 20a \\
\Leftrightarrow -13^{3} + 3 \cdot 13^{2} \sqrt{25277} - 3 \cdot 13 \cdot 25277 + 25277 \sqrt{25277} \gtreqless 20^{3} \cdot 389 \\
\Leftrightarrow -988000 + 25784 \sqrt{25277} \gtreqless 3112000 \Leftrightarrow 25784 \sqrt{25277} \gtreqless 4100000 \\
\Leftrightarrow 3223 \sqrt{25277} \gtreqless 512500
\end{gathered}
\]

Since both sides are positive, we can square them, and the relation is equivalent to

\[
262570625933 \gtreqless 262656250000
\]

This statement is only true for " \(<\) ", so \(z<13\).

Remark: The problem statement is no longer relevant today, as no one would look up cube roots in a four-digit table; instead, they would use a calculator, which is much more accurate and space-saving. Even with just six decimal places of precision, \(\sqrt[3]{636000}-\sqrt[3]{389000} \approx\) \(85.9975-72.9989=12.9986<13\). Such problems are interesting when both sides are equal (as in Problem 041116), because exact equality cannot be shown by arbitrarily precise floating-point arithmetic.