Problem 5. The fifth power of a natural number $n$ consists of the digits $1,2,3,3,7,9$. Find this number.
Solution. Let's form several six-digit numbers (including the smallest and largest) from the given digits and extract their fifth roots using a microcalculator: $\sqrt[5]{123379} \approx 10.42914$; $\sqrt[5]{173923} \approx 11.17047$; $\sqrt[5]{213379} \approx 11.63672$; $\sqrt[5]{337912} \approx 12.75735$; $\sqrt[5]{733921} \approx 14.89807$; $\sqrt[5]{973321} \approx 15.76345$. From this, it is clear that $11 \leqslant n < 15$. But the number $n$ cannot be 15 because the number $15^5$ ends in the digit 5, which is not among the given digits. Therefore, $11 \leqslant n \leqslant 14$.

The number $14^5$ ends in the digit 4, which is also not among the given digits. Therefore, $11 \leqslant n \leqslant 13$.

The number $11^5$ starts and ends with the digit 1, and among the given digits, there is only one 1. Therefore, $n=12$ or $n=13$.

By calculating, we get:

$$
12^5 = 248832, 13^5 = 371293
$$

Thus, $n=13$.