Elementary Number Theory - Divisibility

Question

If $a \mid r$ and $b \mid r$ show that $\operatorname{lcm}(a,b) \mid r$ by using only elementary properties of numbers.

Answer 1

Let $m=\mbox{lcm}(a,b).$ By division algorithm $r=qm+s$ with $0\leq s < m.$ Then $a \mid r = qm+s$. Since $a$ also divides $m$, we have $a \mid s$. Similarly $b\mid s.$ Therefore $s$ is a common multiple of $a$ and $b$. But $m$ is the least common multiple and $s$ is less than $m$. So $s=0$.

Answer 2

$\text{lcm}(a,b)=q$. By division algorithm, $r=mq+c\quad 0\leq c<q.$ We have $a|r,a|q\implies a|c$ and $b|r,b|q\implies b|c$. Since $c$ is a common multiple of $a$ and $b$ with $\:c<q$. So $c=0$ and $q|r$.

Answer 3

Consider the following theorem:

If $m=\operatorname{lcm}(a,b)$ then $m|r$ for any common multiple $r$ of $a,b$.

You can, roughly speaking, say that, since $a|r,b|r$, then $r$ is a common multiple of $a,b$. But each common multiple of $a,b$ is a multiple of $\operatorname{lcm}(a,b)$. Therefore $r=q\operatorname{lcm}(a,b)$. And the proof completes.

The proof of the theorem can be inferred from @B. Goddard answer.