The Minimal number with 2018 Divisors

Relevant wiki: Euler's Theorem

For a positive integer $\displaystyle n = \prod_{k=1}^m p_k^{q_k}$ , where $p_k$ are primes and $q_k$ are the respective powers, then the number of factors or divisors are given by $\displaystyle \sigma_0 (n) = \prod_{k=1}^m (q_k+1)$ . Then for a number $n$ which has 2018 divisors, we have two cases:

$\sigma_0 (n) = 2018 = \begin{cases} 2017 + 1 \\ 2 \times 1009 & \small \color{#3D99F6} \text{Since 2018 has only two prime factors.} \end{cases}$

Case 1: $\sigma_0 (n) = 2018 = 2017+1 = q_1 + 1 \implies n = p_1^{2017}$ . The smallest $n$ in this case is therefore, when $p_1 = 2$ and $n = 2^{2017}$

Case 2: $\sigma_0 (n) = 2018 = 2 \times 1009 = (q_1 + 1)(q_2 + 1) = (1+1)(1008+1) \implies n = p_1^1 \times p_2^{1008}$ . The smallest $n$ in this case is therefore when $p_1 = 3$ and $p_2 = 2$ , that is $n=3\times 2^{1008}$ .

Therefore the smallest $n = 3 \times 2^{1008}$ and we need to find $3\times 2^{1008} \bmod 2018$ . Since $\gcd(2,2018) \ne 1$ , we have to consider the prime factors of 2018 that is 2 and 1009 separately using the Chinese remainder theorem .

Consider prime factor 2: $n = 3\times 2^{1008} \equiv 0 \text{ (mod 2)}$ .

Consider prime factor 1009:

$\begin{aligned} n & \equiv 3\times 2^{\color{#3D99F6} 1008 \bmod \phi (1009)} \text{ (mod 1009)} & \small \color{#3D99F6} \text{Since }\gcd (2,1009) = 1 \text{, Euler's theorem applies.} \\ & \equiv 3\times 2^{\color{#3D99F6} 1008 \bmod 1008} \text{ (mod 1009)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (1009) = 1008 \\ & \equiv 3\times 2^{\color{#3D99F6}0} \equiv 3 \text{ (mod 1009)} \\ \implies n & \equiv 1009 r + 3 & \small \color{#3D99F6} \text{where } r \in \mathbb N \\ 1009 r + 3 & \equiv 0 \text{ (mod 2)} & \small \color{#3D99F6} \text{Since } n \equiv 0 \text{ (mod 2)} \\ \implies r & \equiv 1 \end{aligned}$

Therefore, $n \equiv 1009+3 \equiv \boxed{1012} \text{ (mod 2018)}$ .

A number $n$ is said to be $ordinary$ if the smallest number with exactly $n$ divisors is $p_{1}^{q_{1}-1} ··· p_{a}^{q_{a}-1}$ where $q_{1} ··· q_{a}$ is the prime factorization of $n$ and q_{1} \geq
··· \geq
q_{a} (and where $p_{k}$ denotes the $k$ th prime)

All $square-free$ numbers are $ordinary$ . Please read this paper

$2018$ is a $square-free$ number. Thus, $2018$ is $ordinary$

The prime factorization of $2018$ is $2 \times 1009$
So, the number $n$ we are searching is $2^{1008} \times 3^1$

8229186103190533024883904376780813905111588953898498105168002982684296072639509420747763549304544190419927088006808167617437275794116971854836155577368505449421748051065009396803385296687901036962268811893539692787390930893362253485439137750587184318390678894863264384763210110709215529090973953077280768

and... $n$ $mod 2018$ $=$ $1012$

Here is a program in $Mathematica$ that works only with $ordinary$ numbers

n = 2018; s = Reverse@Flatten[Table @@@ FactorInteger[n]] - 1; Product[Prime@i^s[[i]], {i, Length@s}]

For example if you input $n=80$ to this program it outputs $18480$ which is $NOT$ the smallest number with $80$ divisors.
The smallest number is $15120$ and it gave the wrong anwer because $80$ is $extraordinary$ (not $ordinary$ ).

Here is also a video introduction to Highly Composite Numbers by Numberphile