1. Suppose the natural number $a$ has 15 positive divisors. How many can the natural number $b$ have if the least common multiple of the numbers $a$ and $b$ has 20 positive divisors?
1. In the homework, we showed that a natural number whose prime factorization is $p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{k}^{a_{k}}$, where $p_{1}, p_{2}, \ldots, p_{k}$ are distinct prime numbers and $a_{1}, a_{2}, \ldots, a_{k}$ are natural numbers, has exactly $\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right)$ positive divisors.

There is a unique factorization of the number 15 into a product of several natural numbers greater than 1, which is $15=3 \cdot 5$. Since the number $a$ has 15 divisors, its prime factorization is of the form

$$
p_{1}^{14} \text { or } p_{2}^{2} p_{3}^{4},
$$

where $p_{1}, p_{2} \neq p_{3}$ are prime numbers. All factorizations of the number 20 into a product of several natural numbers greater than 1 are $20=2 \cdot 10=4 \cdot 5=2 \cdot 2 \cdot 5$. Since the least common multiple $n$ of both numbers $a$ and $b$ has 20 divisors, its prime factorization is of the form

$$
q_{1}^{19}, \quad q_{2} q_{3}^{9}, \quad q_{4}^{3} q_{5}^{4} \quad \text { or } \quad q_{6} q_{7} q_{8}^{4}
$$

where $q_{1}, q_{2} \neq q_{3}, q_{4} \neq q_{5}, q_{6} \neq q_{7} \neq q_{8} \neq q_{6}$ are prime numbers.

The number $a$ is also a divisor of the number $n$. This occurs only in the following cases:
a) $p_{1}=q_{1}$, i.e., $a=p_{1}^{14}$ and $n=p_{1}^{19}$. Only $b=p_{1}^{19}$ satisfies this, and the number $b$ thus has exactly 20 positive divisors.
b) $p_{2}=q_{4}$ and $p_{3}=q_{5}$, i.e., $a=p_{2}^{2} p_{3}^{4}$ and $n=p_{2}^{3} p_{3}^{4}$, hence $b=p_{2}^{3} p_{3}^{\alpha}$, where $\alpha \in\{0,1,2,3,4\}$. The number $b$ has $4(1+\alpha)$ divisors in this case, which are all the numbers in the set $\{4,8,12,16,20\}$.

Conclusion. The number $b$ can have $4,8,12,16$ or 20 divisors.

For a complete solution, award 6 points. Of these, award 1 point for listing all possible prime factorizations of the numbers $a$ and $n$, 1 point for finding all cases where $a \mid n$, 1 point for finding all possible forms of the number $b$ in cases a) and b), and 1 point for determining all possible numbers of divisors of the number $b$. A solution that omits one of the cases a) or b) should be awarded no more than 2 points.