Multiples of $a$ are
$$
a, 2a, 3a, 4a, \ldots\,.
$$
Multiples of $b$ are
$$
b, 2b, 3b, 4b, \ldots\,.
$$
First we prove existence of a common multiple: $ab$ is a common multiple: it is the $b$th element of the first list and the $a$th element of the second.
It follows that the set of all common multiples is non-empty.
The well-ordering principle says every non-empty set of positive integers has a least member.
Now we prove uniqueness: There cannot be more than one least member since the integers are linearly ordered.
Q.E.D.
In some cases $ab$ is the least one; in other cases there are smaller common multiples than $ab$.
Pedagogical note:
Show a class of students this:
Multiples of $63$: ${}\quad63,126,189,252,315,378,\ldots$
Multuples of $77$: ${}\quad77,154,231,308,385,\ldots$
Then ask them to vote "yes" or "no" on this question: Is it possible that for either this pair of numbers or some other, the two lists go on forever without having any numbers in common? Have them write their votes on paper and turn them in anonymously and see what you get.
PS: OK, it said divisibly least, so let's add something: All multiples of the one that is numerically least -- call it $c$ -- are common multiples of $a$ and $b$ (proving that part is trivial), and here is the "hard" part: There are no others. For, suppose there were another, and call it $d$. Then the remainder on division of $d$ by $c$ would not be $0$ (since $d$ was assumed not a multiple of $c$) and would be less than $c$. And that remainder would be a common multiple of $a$ and $b$. So we have a common multiple of $a$ and $b$ that is less than $c$, thus a contradiction.
H/T @Bill Dubuque
Set $c=\gcd(a,b)$ and $d=\gcd(a,b-a)$ for convenience.
Step 1: Prove that $c$ is a common divisor of $a,b-a$. This proves $c\le d$.
Step 2: Prove that $d$ is a common divisor of $a,b$. This proves $d\le c$.
Best Answer
Let the numbers be $a,b$. The product of the GCD$(a,b)$ and LCM$(a,b)$ is $ab$. To have something be a factor of the GCD you have to have it be a factor of both $a,b$, so its square must be a factor of $200$. The only numbers whose square is a factor of $200$ are $1,2,5,10$, so those are the only GCDs possible.