Let's find a natural number - preferably a small one - whose cube starts with at least 4 eights (in the decimal system).
The number $N=8888$ does not work, because its prime factorization is $2^{3} \cdot 11 \cdot 101$, so it is not a cube; the cube of the sought number must have at least 5 digits.

Our procedure will be as follows. We will calculate the cube roots of $N$, its tenfold, and its hundredfold to 4 significant digits, rounding down. Using logarithm tables,

$$
r_{1}=\sqrt[3]{8888}=20.71, \quad r_{2}=\sqrt[3]{888880}=44.62, \quad r_{3}=\sqrt[3]{888800}=96.14
$$

(We do not calculate the cube root of 1000 times $N$, as this is $10 r_{1}$, and the digits after the decimal point are the same as those of $r_{1}$.)

From these, ignoring the decimal point and raising to the cube, the number formed by the first four digits of the cube is smaller than $N$, but if we round up to 2, 3, or 4 significant digits, the number is larger than $N$. We will search until we find a cube number whose first four digits are smaller than $N+1$. Now,

$$
\begin{aligned}
& r_{1} \text { gives } \quad 21^{3}=9261 \quad(4 \text { digits }) \\
& r_{2} \text { gives } \quad 45^{3}=91 \ldots \quad(5 \text { digits }) \\
& r_{3} \text { gives } \quad 97^{3}=91 \ldots \quad(6 \text { digits }) \\
& r_{1} \text { gives } \quad 208^{3}=899 \ldots \quad(7 \text { digits }) \\
& r_{2} \text { gives } \quad 447^{3}=893 \ldots \quad(8 \text { digits }) \\
& r_{3} \text { gives } \quad 962^{3}=890 \ldots \quad(9 \text { digits }) \\
& r_{1} \text { gives } \quad 2072^{3}=8895 \ldots \quad(10 \text { digits }) \\
& r_{2} \text { gives } \quad 4463^{3}=88895 \ldots \quad(11 \text { digits }) \\
& r_{3} \text { gives } \quad 9615^{3}=8888896 \ldots(12 \text { digits })
\end{aligned}
$$

Accordingly, the smallest natural number that starts with 8888 when cubed is 9615, and its cube starts with 5 eights.

The raising to the cube of 3 and 4 digit bases could not have been done with reassuring accuracy using logarithms. In the repeated multiplications, we omitted the trailing digits, leaving only as many at the front as were needed to determine the number of leading 8s.

Csetényi Artúr (Szeged, Radnóti M. Gymnasium, III. grade)