3. Determine all natural numbers $M$ divisible by 240, for which the equation $M = \operatorname{LCM}(x, y)$ with unknowns $x$ and $y$ has exactly 1001 solutions in the domain of natural numbers. (The symbol $\mathrm{LCM}(x, y)$ denotes the least common multiple of the numbers $x$ and $y$.)
SOLUTION. First, let's show that for a number $M$ with prime factorization $M=\prod_{i=1}^{n} p_{i}^{c_{i}}$ (where $p_{i}$ are distinct primes), the number of solutions of the equation $\mathrm{LCM}(x, y)=M$ is equal to $\prod_{i=1}^{n}\left(2 c_{i}+1\right)$. Indeed, each solution $(x, y)$ of the given equation has the property that any prime $p_{i}(i=1, \ldots, n)$ divides at least one of the numbers $x$ and $y$ (and at most $\mathrm{v}$ such powers, $\mathrm{v}$ which divides $M$) and no other primes divide either $x$ or $y$; $x$ and $y$ are thus of the form $x=\prod_{i=1}^{n} p_{i}^{a_{i}}, y=\prod_{i=1}^{n} p_{i}^{b_{i}}$, $a_{i}, b_{i} \in \mathbb{N}_{0}$, and moreover $\max \left(a_{i}, b_{i}\right)=c_{i}, i=1, \ldots, n$. The numbers $x$ and $y$ thus uniquely determine $n$-tuples of numbers $a_{i}$ and $b_{i}$ and are uniquely determined by them. All solutions of the given equation are thus described by pairs of $n$-tuples of natural numbers such that at the $i$-th position there is a number from the set $\left\{0, \ldots, c_{i}\right\}$ in both $n$-tuples and at least in one of them it is exactly equal to $c_{i}$. There are $\prod_{i=1}^{n}\left(2 c_{i}+1\right)$ such $n$-tuples: Two $n$-tuples of numbers $\left(a_{1}, a_{2}, \ldots, a_{n}\right)$ and $\left(b_{1}, b_{2}, \ldots, b_{n}\right)$ can be considered as $n$ pairs of numbers $\left(a_{1}, b_{1}\right),\left(a_{2}, b_{2}\right), \ldots,\left(a_{n}, b_{n}\right)$. Any pair $\left(a_{i}, b_{i}\right)$ can independently take $\left(2 c_{i}+1\right)$ different values $\left(0, c_{i}\right),\left(1, c_{i}\right), \ldots,\left(c_{i}-1, c_{i}\right),\left(c_{i}, c_{i}\right),\left(c_{i}, c_{i}-1\right), \ldots$, $\left(c_{i}, 1\right),\left(c_{i}, 0\right)$. According to the combinatorial rule of product, we get the above number of solutions.

The prime factorization of the number 1001 is 7$\cdot$11$\cdot$13. For the given equation to have exactly 1001 solutions, the exponents $c_{i}$ from the prime factorization of the number $M$ (containing at least three primes, namely 2, 3, and 5) must satisfy the equation $\prod_{i=1}^{n}\left(2 c_{i}+1\right)=7 \cdot 11 \cdot 13$. Therefore, in the prime factorization of the number $M$, there must be exactly three primes, and they must be in the powers $\frac{1}{2}(7-1)=3$, $\frac{1}{2}(11-1)=5$ and $\frac{1}{2}(13-1)=6$. Since $M$ must be divisible by the number $240=2^{4} \cdot 3 \cdot 5$, i.e., by the primes 2, 3, and 5 in the corresponding powers, the only possible choices for $M$ are the numbers $2^{5} \cdot 3^{3} \cdot 5^{6}, 2^{5} \cdot 3^{6} \cdot 5^{3}, 2^{6} \cdot 3^{5} \cdot 5^{3}, 2^{6} \cdot 3^{3} \cdot 5^{5}$.

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