Why does prime factorization for LCM work?

Question

A method to finding the LCM of two numbers is to use prime factoring. I know how to do this, but logically why does this work in finding the lcm

Answer 1

By FTA (existence and uniqueness of prime factorizations), divisibility reduces to divisibility in each prime component, i.e. $\ p^{\large b} q^{\large b'}\!\cdots\mid p^{\large a} q^{\large a'}\!\cdots\!\iff p^{\large b}\mid p^{\large a}\ $ & $\,\ q^{\large b'}\!\mid q^{\large a'}\ \ldots\ $

So $\,\ B,C\mid A\iff p^{\large b},p^{\large c}\mid p^{\large a}\ $ & $\,\ q^{\large b'},q^{\large c'}\mid q^{\large a'}\ \ldots\ $ But we have

$$\qquad\qquad\ \, p^{\large b},p^{\large c}\mid p^{\large a}\! \iff b,c\le a \iff \max\{b,c\}\le a \iff p^{\large \max\{b,c\}}\mid p^{\large a}$$

Reassembling the prime components yields the result for the lcm

$$\quad B,C\mid A \iff \color{#c00}{p^{\max\{b,c\}} q^{\max\{b',c'\}}}\ldots\mid A$$

Remark $\ $ Above we employ the universal characterization of lcm, i.e.

$$\begin{align} \ \ \ \ \ B,C\mid A\iff \color{#c00}{{\rm lcm}(B,C)}\mid A\end{align}\qquad\qquad$$

Answer 2

The easiest way for the lcm is to find the gcd first by Euclid's algorithm (no prime factorisation needed) and then the lcm of $n$ and $m$ is just $\frac{nm}{\gcd(n,m)}$. Read the enlightening Wikipedia page.