1. Find a natural number that has six natural divisors (including one and the number itself), two of which are prime, and the sum of all its natural divisors is 78.
Solution: The desired natural number $n$ can be represented as $n=p_{1}^{\alpha_{1}} \cdot p_{2}^{\alpha_{2}}, 12,1+p_{2}+p_{2}^{2}>7$, then no factorization of the number 78 into a product of two natural factors fits $78=1 \cdot 78,78=2 \cdot 39,78=3 \cdot 26,78=6 \cdot 13$ up to the order).
2) $\alpha_{1}=2, \alpha_{2}=1, n=p_{1}^{2} \cdot p_{2}$. Since the sum of all divisors is 78, then $\left(1+p_{1}+p_{1}^{2}\right)\left(1+p_{2}\right)=78$. Since $13,1+p_{1}+p_{1}^{2}>3$, we have

a) $1+p_{1}+p_{1}^{2}=6,1+p_{2}=13$, there are no natural solutions.

b) $1+p_{1}+p_{1}^{2}=13,1+p_{2}=6$, or $p_{1}=3, p_{2}=5, n=3^{2} \cdot 5=45$.

Answer: 45.