1. Determine all natural numbers $n$ such that using each of the digits $0,1,2,3, \ldots, 9$ exactly once, the numbers $n^{3}$ and $n^{4}$ can be written.
Solution. If $n$ is a one-digit number, then $n^{3} \leq 9^{3}=729, n^{4} \leq 9^{4}=6561$, i.e., the numbers $n^{3}$ and $n^{4}$ have fewer than 7 digits. If the number $n$ has more than two digits, then $n^{3} \geq 100^{3}=10^{6}, n^{4} \geq 100^{4}=10^{8}$, so the numbers $n^{3}$ and $n^{4}$ have more than $7+9=16$ digits. Let each digit $0,1,2, \ldots, 9$ appear exactly once in the records of the numbers $n^{3}$ and $n^{4}$. From the above, it follows that the number $n$ is a two-digit number. Furthermore, it easily follows that $n^{3}$ is a four-digit number, and $n^{4}$ is a six-digit number. From the condition $1000 \leq n^{3} \leq 9999$ it follows that $10 \leq n \leq 21$, and from the condition $100000 \leq n^{4} \leq 999999$ it follows that $18 \leq n \leq 31$. Therefore, $n \in\{18,19,20,21\}$. Each of the numbers $20^{3}$ and $20^{4}$ ends in 0, and each of the numbers $21^{3}$ and $21^{4}$ ends in 1. Furthermore, the number $19^{4}=130321$ has repeating digits, so it remains to check whether $n=18$ is a solution to the problem. We have. $18^{3}=5832$ and $18^{4}=104876$, i.e., the number 18 satisfies the conditions of the problem and is its only solution.