Is it possible for three numbers to not have an LCM?

Question

For a software application, I need to generate three very large sequences of numbers (say $n \gt 100000$) such that there are no common values. I think this problem boils down to finding three such numbers e.g. $x$, $y$, $z$, such that there is no LCM for those three numbers. Do any such three numbers exist? $x$, $y$ and $z$ also have the following constraints: $0 \leq x \leq 30$, $31 \leq y \leq 60$, and $61 \leq z \leq 91$.

Answer 1

To answer your question, no, there are no three numbers whose LCM does not exist.

Given the three numbers, $x$, $y$ and $z$, there exists a multiple $xyz$, regardless of any restrictions. This means the set of common multiples is nonempty. Since this set is nonempty, there exists a least element.

Please have a read through this source for more information, specifically Noah Schweber's answer.

I hope you found this helpful!