There exists an integer $m$ such that $1\vert m$, $2\vert m$, $3\vert m$, $4\vert m$,... $n\vert m$.

Question

So, the exact question is, given that $n > 1$ is an integer, prove that there exists an integer $m$ such that $2\vert m$, $3\vert m$, $4\vert m$,... $n\vert m$. I am beyond lost on this, so any sort of direction would be greatly appreciated.

Given that it is referring to divisibility I assume that I may have to reference that an integer $a$ is said to be divisible by $b$ if there exists an integer $c$ such that $a = b \cdot c$, but I'm not entirely sure how I could generalize this form for my problem.

Answer 1

Hint: Consider $lcm(1,2,\ldots,n)$, defined as the least common multiplier of $1,2,\ldots,n$.