1. We will call a prime number nice if it can be written as the difference of two cubes of natural numbers. Determine the last digit of all nice prime numbers.
1. First, notice that $5^{3}-4^{3}=61$, $2^{3}-1^{3}=7$, and $3^{3}-2^{3}=19$ are all nice prime numbers, so 1, 7, and 9 belong to the sought-after last digits. We will show that no other digits can be the last digit.

Choose a nice prime number $p$ and write it as $m^{3}-n^{3}$, where $m>n$ are natural numbers. According to the well-known formula, we have

$$
p=m^{3}-n^{3}=(m-n)\left(m^{2}+m n+n^{2}\right),
$$

and since the second parenthesis is greater than 1, it means that $m=n+1$ (otherwise, we would factorize the prime number $p$ into a product of two factors greater than 1). Substituting this, we get

$$
p=3 n^{2}+3 n+1.
$$

Since $3 n^{2}+3 n+1>6$, the prime number $p$ must be odd and different from five. This eliminates $0, 2, 4, 5, 6$, and 8 as possible last digits, and only the digit 3 remains to be excluded.

To do this, it suffices to determine the remainders when numbers of the form $3 n^{2}+3 n+1$ are divided by five. Substituting the remainders $0, 1, 2, 3$, and 4 into (1) successively yields the remainders $1, 2, 4, 2, 1$, thus excluding the remainder 3.

Answer: The last digits of nice prime numbers are precisely 1, 7, and 9.

Note. Deriving relation (1) is not necessary. Once the insight $m=n+1$ is obtained, it suffices to find the possible last digits of the differences of the cubes of two consecutive natural numbers. To do this, we first create a table of the last digits of numbers $k$ and $k^{3}$:

$$
\begin{array}{ccccccccccc}
k & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\
k^{3} & 0 & 1 & 8 & 7 & 4 & 5 & 6 & 3 & 2 & 9
\end{array}
$$

We see that the last digits of the differences $(k+1)^{3}-k^{3}$ are

$$
1-0, 8-1, 17-8, 14-7, 5-4, 6-5, 13-6, 12-3, 9-2, 10-9
$$

thus only the digits 1, 7, and 9. (However, it is still necessary to provide examples of nice prime numbers ending in these last digits.)

For a complete solution, award 6 points. For finding nice prime numbers ending in each of the digits 1, 7, and 9, award 1 point in total. Award 1 point for any correct argument leading to the exclusion of each of the digits $0, 2, 4, 5, 6$, and 8. Award another 2 points for deriving that the sought-after primes have the form $3 n^{2}+3 n+1$. The remaining 2 points are for completing the solution. In the case of the approach from the note, deduct 1 point if the solver omits providing corresponding examples of nice prime numbers.