How many natural numbers are there, the square and cube of which, when written together, require 10 digits?
A larger natural number has a larger square and cube, and to write the larger of two different natural numbers, the same number or more digits are needed than to write the smaller one.

The square and cube of the largest single-digit number together use only 5 digits: $9^{2}=81,9^{3}=729$, so the numbers in question must be at least two-digit. However, in the case of the smallest three-digit number, 12 digits are used: $100^{2}=10000,100^{3}=1000000$, so the numbers in question are all two-digit.

For every two-digit number $n$, the number of digits in its square is: $a=3$, or 4, because $10^{2}=100$ and $99^{2}=9801$. The change occurs when $n^{2}$ exceeds the smallest four-digit number, 1000. This is not a perfect square, its square root (to one decimal place) is 31.6, so $31^{2}$ is still three-digit, and $32^{2}$ is already four-digit.

In the cube of two-digit numbers, the number of digits is $\beta=4,5$, or 6, because $10^{3}=1000$ and $99^{3}=970299$. Here, the change occurs when $n^{3}$ exceeds 10000 or 100000. According to the table, the cube root (to one decimal place) of these is 21.5 and 46.4, respectively, so the five-digit cubes start with $22^{3}$, and the six-digit ones with $47^{3}$.

We only get 10 from the sum of $\alpha$ and $\beta$ if $\alpha=4$ and $\beta=6$. According to this, $n \geqq 32$ and $n \geqq 47$, so the smallest suitable number is 47, the largest is 99, and their count is $99-46=53$.

László Fazekas (Szigetvár, Zrínyi M. g. II. o. t.),

Note. The "slightly flawed" solutions correctly determined the above limits for $n$, but gave 52 numbers as the final result.