Divisible by the numbers from 1 up to 20

First of all, let's prove that, in fact, exists such number. Let $X= \{ n \in \mathbb{N}: \forall i\in \{1,\dotsc , 20\}, i|n \}$ .

Let's consider the number $20!$ . In these conditions, it is obvious that $\forall i\in \{1,\dotsc , 20\}, i|(20!)$ . Therefore, $20! \in X$ .

As $X \subseteq \mathbb{N}$ and $X \neq \emptyset$ we have that $X$ has a minimum in $\mathbb{N}$ .

Let $p= \min X$ .

Let's analyse what are the necessary factores that $p$ should have for being that smallest element in $X$ .

It is clear that $2^4$ should be one of those factores, because $p|16$ . It is also clear that all prime numbers in $\{1,\dotsc , 20\}$ should be factores of $p$ , i.e., $3,5,7,11,13,17$ and $19$ are factores of $p$ . We also know that $p|3^2$ , so 3^2 should be one of the factores of $p$ .

Let's see if $p= 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ . Let $m=2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19$ . It is clear that $\forall s\in \{2,3,5,7,11,13,17,19\} , s|m$ .

And it is also clear that

$4|m$

$6|m$

$8|m$

$9|m$

$10|m$

$12|m$

$14|m$

$15|m$

$16|m$

$20|m$ .

So, by construction, we can conclude that $p=m=2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19=232792560$ .