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

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 $s=0$.
