5. Find the sum of all natural numbers that have exactly four natural divisors, three of which (of the divisors) are less than 15, and the fourth is not less than 15.
Answer: $\left((2+3+5+7+11+13)^{2}-\left(2^{2}+3^{2}+5^{2}+7^{2}+11^{2}+13^{2}\right)\right) / 2-$ $6-10-14+27=649$.

Solution: The numbers $N$ specified have exactly 4 divisors either if $N=$ $p^{3}$, or if $N=p q$, where $p$ and $q$ are primes. In the first case, only 27 fits. In the second case, we need to consider the pairwise products of all prime numbers less than 15. These numbers are $p_{i}=2,3,5,7,11,13$. The sum of their pairwise products can be calculated as $S_{p q}=2 \cdot(3+5+7+$ $11+13)+3 \cdot(5+7+11+13)+5 \cdot(7+11+13)+7 \cdot(11+13)+11 \cdot 13=652$. From the pairwise products, we need to exclude 6, 10, and 14, as they are less than 15. In the end, $S=27+S_{p q}-6-10-14=649$.