4. Let $m$ be a natural number that has 7 positive divisors, and $n$ be a natural number that has 9 positive divisors. How many divisors can the product $m \cdot n$ have?
SOLUTION. Let $r=p_{1}^{a_{1}} p_{2}^{a_{2}} \ldots p_{k}^{a_{k}}$ be the prime factorization of a natural number $r$, where $p_{1}, p_{2}, \ldots, p_{k}$ are distinct prime numbers and $a_{1}, a_{2}, \ldots, a_{k}$ are positive integers. (Such a factorization is unique up to the order of the primes.) Each divisor $d$ of the natural number $r$ is then of the form $d=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \ldots p_{k}^{\alpha_{k}}$, where

$$
0 \leqq \alpha_{1} \leqq a_{1}, 0 \leqq \alpha_{2} \leqq a_{2}, \ldots, 0 \leqq \alpha_{k} \leqq a_{k}
$$

The number of divisors thus corresponds exactly to the number of ways to choose a $k$-tuple of non-negative integers $\left(\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k}\right)$ satisfying conditions (1). Since each of the numbers $\alpha_{i}$ can be chosen in exactly $a_{i}+1$ ways $(1 \leqq i \leqq k)$, the number of divisors of the natural number $r$ is given by

$$
\tau(r)=\left(a_{1}+1\right)\left(a_{2}+1\right) \ldots\left(a_{k}+1\right) .
$$

Note that each of the factors in (2) is greater than 1. Since the number $m$ has 7 divisors (the number 7 is a prime number), $m$ has a prime factorization of the form $m=p^{6}$. Similarly, all numbers $n$ with 9 divisors have a prime factorization of the form $n=q_{1}^{8}$ or $n=q_{1}^{2} q_{2}^{2}$, where $q_{1} \neq q_{2}$. Indeed, $3 \cdot 3$ is the only factorization of the number 9 into factors greater than 1.

Let's now consider all possibilities, keeping in mind that the prime number $p$ can be equal to one of the prime numbers $q_{i}$.

A. Case $n=q_{1}^{8}$.

a) Suppose $p \neq q_{1}$. Then $m n=p^{6} q_{1}^{8}$ is the prime factorization of the number $m n$. The number $m n$ has $7 \cdot 9=63$ divisors in this case.

b) Suppose $p=q_{1}$. Then $m n=p^{6} p^{8}=p^{14}$ is the prime factorization of the number $m n$. The number $m n$ has 15 divisors in this case.

B. Case $n=q_{1}^{2} q_{2}^{2}$.

a) Suppose $q_{1} \neq p \neq q_{2}$. Then $m n=p^{6} q_{1}^{2} q_{2}^{2}$ is the prime factorization of the number $m n$. The number $m n$ has $7 \cdot 3 \cdot 3=63$ divisors in this case.

b) One of the primes $q_{1}, q_{2}$ is equal to $p$. Since the primes in the factorization of $n$ are in the same powers, it suffices to consider one of these cases, for example, $p=q_{1}$. Then $m n=p^{6} p^{2} q_{2}^{2}=p^{8} q_{3}^{2}$ is the prime factorization of the number $m n$. The number $m n$ has $9 \cdot 3=27$ divisors in this case.

The number of divisors of the number $m n$ can be 15, 27, or 63. The appropriate number of divisors can be obtained, for example, by taking $m=64=2^{6}$ and the numbers $n$ equal to $256=2^{8}, 100=2^{2} 5^{2}$, or $225=3^{2} 5^{2}$.

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