Can't understand a proof: Let $a,b,c$ be integers. If $a$ and $b$ divide $c$, then $lcm(a,b)$ also divides $c$ Can't understand a proof: Let $a,b,c$ be integers. If $a$ and $b$ divide $c$, then $lcm(a,b)$ also divides $c$

Proof I've been shown: 
Since $c$ is a common multiple of $a,b$, it means that $lcm(a,b) \leq c$.

So:
$c=q\cdot lcm(a,b)+r$, $q \in \mathbb Z$ and $0\leq r \leq lcm(a,b)$

Since $c$ and $lcm(a,b)$ are common multiples of $a,b$, then so is $r$. But $m$ is the least common divisor, then it means $r=0$. 

[Proof taken from: http://www.math.utah.edu/~wood/4400/hw1sol.pdf]

I cannot understand the last step... Why does it imply that $r=0$? 

Thank you!
It follows from the (general) fact that if $a$ divides $b$ and $c$, then $a$ also divides $b-c$, so in this case:

$c = q.lcm(a,b) + r$ and $0 \leq r &lt; lcm(a,b)$

$c - q.lcm(a,b) = r$

Which means that both $a$ and $b$ divides $r$, but $0 \leq r &lt; lcm(a,b)$. So $r$ must be equal to $0$, and we get:

$c = q.lcm(a,b)$, as you wish.
I think it's easier using examples. if a=1, b=5, lcm=5.
If we pick c, a common multiple of 1 and 5, such as c=10 for example,
then c=q.lcm : in this case 10 = 2 * 5 + r
since lcm is the least common divisor, it means it is a direct multiple, so r=0 as seen from the example.