IRL ABooklet

The divisors of an integer $n$ with prime factorisation
$$n = p_1^{e_1} \cdot p_2^{e_2} \cdots p_k^{e_k}$$
are all of the form $p_1^{f_1} \cdot p_2^{f_2} \cdots p_k^{f_k}$, where $0 \le f_i \le e_i$. Hence, the number of positive divisors of $n$ is equal to $(e_1+1)(e_2+1)\cdots(e_k+1)$ and for each prime number $p_i$ that divides $n$, we have $e_i+1 \ge 2$.
We are looking for the smallest $n$ for which $(e_1+1)(e_2+1)\cdots(e_k+1) = 2020$. Because $2020 = 2^2 \cdot 5 \cdot 101$, $n$ can have at most $4$ different prime divisors. We will consider, one after the other, the cases $k=1, 2, 3, 4$.

k = 4. The four factors $e_i+1$ must be $2, 2, 5, 101$, hence the $e_i$ are $1, 1, 4, 100$. To make the number as small as possible, we pair the big $e_i$ with the smallest primes possible. This gives
$$n = 2^{100} \cdot 3^4 \cdot 5^1 \cdot 7^1 < 2^{100+8+3+3} = 2^{114}.$$

k = 3. The set of three factors $e_i+1$ must be either $\{4, 5, 101\}$, or $\{2, 10, 101\}$, or $\{2, 5, 202\}$, or $\{2, 2, 505\}$. Again, pairing the larger powers with the smaller primes, we get the following possibilities:
$$2^{100} \cdot 3^4 \cdot 5^3 > 2^{100+4+6} = 2^{110}$$
$$2^{100} \cdot 3^9 \cdot 5^1 > 2^{100+9+2} = 2^{111}$$
$$2^{201} \cdot 3^4 \cdot 5^1 > 2^{201}$$
$$2^{504} \cdot 3^1 \cdot 5^1 > 2^{504}$$
We can reject the last two candidates as being too big. We will return to the first two at the end.

k = 2. With two factors, we are only interested in case where the result will be less than $2^{114}$, so both exponents, $e_1$ and $e_2$, must be less than $57$, otherwise result will be at least $2^{57} \cdot 3^{57} > 2^{114}$. One of the two factors $e_i+1$ of $2020$ must be larger than $\sqrt{2020} > 44$, so we are looking for a factor of $2020$ between $44$ and $58$. From the prime factorisation of $2020$ we see that there is no factor in this range.

k = 1. The smallest possibility with one factor is $2^{2019}$, which is way too big. We are left with three candidates:
$$2^{100} \cdot 3^4 \cdot 5 \cdot 7,$$
$$2^{100} \cdot 3^4 \cdot 5^3$$
$$2^{100} \cdot 3^9 \cdot 5$$
and need to decide which is the smallest. We can take out a common factor of $2^{100} \cdot 3^4 \cdot 5$ and compare $7, 5^2, 3^5$ and find the first is the smallest. Thus $2^{100} \cdot 3^4 \cdot 5 \cdot 7$ is the smallest number with exactly $2020$ distinct positive divisors.