17.1. [9.6 (15 points), 10.3 (15 points)] Find the prime factorization of the smallest natural number that has exactly 2020 distinct natural divisors.
Answer: $2^{100} \cdot 3^{4} \cdot 5 \cdot 7$.

Solution. The number of $\tau(n)$ different natural divisors of a number $n$, which admits a factorization into prime factors

$$
n=p_{1}^{\alpha_{1}} p_{2}^{\alpha_{2}} \cdot \ldots \cdot p_{k}^{\alpha_{k}}
$$

(here $\alpha_{1}, \alpha_{2}, \ldots, \alpha_{k} \geqslant 1$), is expressed by the formula

$$
\tau(n)=\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdot \ldots \cdot\left(\alpha_{k}+1\right)
$$

Indeed, any divisor of the number $n$ has the form $p_{1}^{\beta_{1}} p_{2}^{\beta_{2}} \cdot \ldots \cdot p_{k}^{\beta_{k}}$, where $0 \leqslant \beta_{i} \leqslant \alpha_{i}, i=1, \ldots, k$, and each exponent $\beta_{i}$ can be chosen in $\alpha_{i}+1$ ways (from 0 to $\alpha_{i}$).

Thus, we need to find the smallest number $n$ of the form (1), for which

$$
\left(\alpha_{1}+1\right)\left(\alpha_{2}+1\right) \cdot \ldots \cdot\left(\alpha_{k}+1\right)=2020=2^{2} \cdot 5 \cdot 101
$$

that is, to determine the corresponding exponents $\alpha_{1}, \ldots, \alpha_{k}$ and prime factors $p_{1}, \ldots, p_{k}$.

We will show that the desired number is $2^{100} \cdot 3^{4} \cdot 5 \cdot 7$.

We can assume (relabeling the prime factors if necessary) that the exponents in the factorization (1) are ordered in non-increasing order: $\alpha_{1} \geqslant \alpha_{2} \geqslant \ldots \geqslant \alpha_{k}$. Moreover, it is not difficult to see that for a fixed (non-increasingly ordered) set of exponents, among all numbers of the form (1), the smallest is the number for which $p_{1}, p_{2}, \ldots, p_{k}$ are the first $k$ prime numbers, taken in increasing order ${ }^{1}$. Indeed, if $p_{i}>p_{j}, \alpha_{i}>\alpha_{j}$, then $p_{i}^{\alpha_{i}} p_{j}^{\alpha_{j}}>p_{j}^{\alpha_{i}} p_{i}^{\alpha_{j}}$ (since $p_{i}^{\alpha_{i}-\alpha_{j}}>p_{j}^{\alpha_{i}-\alpha_{j}}$), and swapping the factors $p_{i}$ and $p_{j}$ allows us to reduce the desired number. Clearly, all the smallest prime factors must be used.

Furthermore, note that the number 101 must enter the product (2) as a separate factor, since otherwise $\alpha_{1}+1 \geqslant 101 \cdot 2$ and

$$
p_{1}^{\alpha_{1}} \geqslant 2^{201}>2^{100} \cdot 3^{4} \cdot 5 \cdot 7
$$

It remains to consider the possible factorizations of the number 2020 with a separate factor of 101:

$$
101 \cdot 20, \quad 101 \cdot 10 \cdot 2, \quad 101 \cdot 5 \cdot 4, \quad 101 \cdot 5 \cdot 2 \cdot 2
$$

and compare the numbers

$$
2^{100} \cdot 3^{19}, \quad 2^{100} \cdot 3^{9} \cdot 5, \quad 2^{100} \cdot 3^{4} \cdot 5^{3}, \quad 2^{100} \cdot 3^{4} \cdot 5 \cdot 7
$$